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How to prove the triangle inequality for this metric space
Show that in a discrete metric space, every subset is both open and closed.Proof that a discrete metric is indeed a metric spaceContinuity of functions in a metric space.What characteristics of binary space are needed to prove that it is totally disconnected?Continuous functions in a metric space using the discrete metricMetric space - complete - discreteThe limit points of a subset of binary space with ultra metricGiven a distance function $d$ prove it is a metric on the Baire space.How does the metric on a space $X$, affect the way we can define neighborhoods of points $ p in X$?How to prove the set of bounded sequences is clopen in the uniform metric?
$begingroup$
Let $X = prod_{k=1}^infty {0,1}$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = min{k : f(k) neq g(k)}$ and define $d : X times X to mathbb{R}$ by
$$d(f, g) = begin{cases}2^{-m(f,g)} & text{if } f neq g \ 0 & text{if }f = g.end{cases}$$
The first three axioms have been given I imagine the first starting point is to prove that
$$m(a,c) ge min{m(a,b), m(b,c)}.$$
Followed by:
For $f in X$, $n in mathbb{N}$, prove that $B(f, 2^{-n})$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text{ and }f(n) = g(n).$$
This I think you must use the previous fact that $d(f,g) = 2^{-m(f,g)}$.
Followed by:
Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.
This I believe must be a continuity using open sets problem.
general-topology metric-spaces
$endgroup$
|
show 11 more comments
$begingroup$
Let $X = prod_{k=1}^infty {0,1}$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = min{k : f(k) neq g(k)}$ and define $d : X times X to mathbb{R}$ by
$$d(f, g) = begin{cases}2^{-m(f,g)} & text{if } f neq g \ 0 & text{if }f = g.end{cases}$$
The first three axioms have been given I imagine the first starting point is to prove that
$$m(a,c) ge min{m(a,b), m(b,c)}.$$
Followed by:
For $f in X$, $n in mathbb{N}$, prove that $B(f, 2^{-n})$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text{ and }f(n) = g(n).$$
This I think you must use the previous fact that $d(f,g) = 2^{-m(f,g)}$.
Followed by:
Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.
This I believe must be a continuity using open sets problem.
general-topology metric-spaces
$endgroup$
6
$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago
1
$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
2 hours ago
1
$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago
2
$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too:neq
= "$neq$",in
= "$in$",ge
= "$ge$". Try putting$
signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
$endgroup$
– Theo Bendit
2 hours ago
2
$begingroup$
$What is B'???$
$endgroup$
– William Elliot
1 hour ago
|
show 11 more comments
$begingroup$
Let $X = prod_{k=1}^infty {0,1}$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = min{k : f(k) neq g(k)}$ and define $d : X times X to mathbb{R}$ by
$$d(f, g) = begin{cases}2^{-m(f,g)} & text{if } f neq g \ 0 & text{if }f = g.end{cases}$$
The first three axioms have been given I imagine the first starting point is to prove that
$$m(a,c) ge min{m(a,b), m(b,c)}.$$
Followed by:
For $f in X$, $n in mathbb{N}$, prove that $B(f, 2^{-n})$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text{ and }f(n) = g(n).$$
This I think you must use the previous fact that $d(f,g) = 2^{-m(f,g)}$.
Followed by:
Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.
This I believe must be a continuity using open sets problem.
general-topology metric-spaces
$endgroup$
Let $X = prod_{k=1}^infty {0,1}$ which can be viewed as the set of all sequences of binary digits. For $f,g in X$, with $f neq g$, define $m(f, g) = min{k : f(k) neq g(k)}$ and define $d : X times X to mathbb{R}$ by
$$d(f, g) = begin{cases}2^{-m(f,g)} & text{if } f neq g \ 0 & text{if }f = g.end{cases}$$
The first three axioms have been given I imagine the first starting point is to prove that
$$m(a,c) ge min{m(a,b), m(b,c)}.$$
Followed by:
For $f in X$, $n in mathbb{N}$, prove that $B(f, 2^{-n})$ consists of all $g in X$ for which
$$f(1) = g(1), f(2) = g(2), ldots, text{ and }f(n) = g(n).$$
This I think you must use the previous fact that $d(f,g) = 2^{-m(f,g)}$.
Followed by:
Show that for all $f in X$, $r > 0$, either $B'(f,r) = B(f,r)$ or $B'(f,r) = B(f,2r)$.
This I believe must be a continuity using open sets problem.
general-topology metric-spaces
general-topology metric-spaces
edited 1 hour ago
Theo Bendit
19.3k12353
19.3k12353
asked 3 hours ago
Alexander QuinnAlexander Quinn
265
265
6
$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago
1
$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
2 hours ago
1
$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago
2
$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too:neq
= "$neq$",in
= "$in$",ge
= "$ge$". Try putting$
signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
$endgroup$
– Theo Bendit
2 hours ago
2
$begingroup$
$What is B'???$
$endgroup$
– William Elliot
1 hour ago
|
show 11 more comments
6
$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago
1
$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
2 hours ago
1
$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago
2
$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too:neq
= "$neq$",in
= "$in$",ge
= "$ge$". Try putting$
signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.
$endgroup$
– Theo Bendit
2 hours ago
2
$begingroup$
$What is B'???$
$endgroup$
– William Elliot
1 hour ago
6
6
$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago
$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago
1
1
$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
2 hours ago
$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
2 hours ago
1
1
$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago
$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago
2
2
$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too:
neq
= "$neq$", in
= "$in$", ge
= "$ge$". Try putting $
signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.$endgroup$
– Theo Bendit
2 hours ago
$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too:
neq
= "$neq$", in
= "$in$", ge
= "$ge$". Try putting $
signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.$endgroup$
– Theo Bendit
2 hours ago
2
2
$begingroup$
$What is B'???$
$endgroup$
– William Elliot
1 hour ago
$begingroup$
$What is B'???$
$endgroup$
– William Elliot
1 hour ago
|
show 11 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$
Suppose $d(f,g)+d(g,h)<d(f,h).$
Then $ 2^{-a}+2^{-b}<2^{-c},$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text { as }c<a=m(f,g),)$$ $$ text {and } quad g(c)=h(c)quad (text { as } c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.
$endgroup$
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
add a comment |
$begingroup$
If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min {m(a, b), m(b, c)} - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
$$m(a, c) ge min{m(a, b), m(b, c)}.$$
Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
begin{align*}
d(a, b) + d(b, c) &= 2^{-m(a, b)} + 2^{-m(b, c)} \
&= 2^{-max{m(a, b), m(b, c)}} + 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-m(a, c)} = d(a, c).
end{align*}
If $g in B(f, 2^{-n})$, then either $f = g$ (in which case, the result is trivially true), or
$$2^{-n} > d(f, g) = 2^{-m(a, b)} implies m(a, b) > n.$$
So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.
Note that the distance function $d(f, g)$ only takes countably many values:
$$0, frac{1}{2}, frac{1}{4}, ldots, frac{1}{2^n}, ldots$$
When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
$$Bleft(f, frac{1}{3}right) = Bleft(f, frac{1}{2}right) = B'left(f, frac{1}{4}right),$$
simply because a point that at most $frac{1}{3}$ distance away from $f$, basically only rules out their distances being $frac{1}{2}$.
This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.
$endgroup$
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
|
show 2 more comments
$begingroup$
For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.
For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.
Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball
$endgroup$
add a comment |
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3 Answers
3
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oldest
votes
3 Answers
3
active
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$begingroup$
Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$
Suppose $d(f,g)+d(g,h)<d(f,h).$
Then $ 2^{-a}+2^{-b}<2^{-c},$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text { as }c<a=m(f,g),)$$ $$ text {and } quad g(c)=h(c)quad (text { as } c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.
$endgroup$
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
add a comment |
$begingroup$
Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$
Suppose $d(f,g)+d(g,h)<d(f,h).$
Then $ 2^{-a}+2^{-b}<2^{-c},$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text { as }c<a=m(f,g),)$$ $$ text {and } quad g(c)=h(c)quad (text { as } c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.
$endgroup$
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
add a comment |
$begingroup$
Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$
Suppose $d(f,g)+d(g,h)<d(f,h).$
Then $ 2^{-a}+2^{-b}<2^{-c},$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text { as }c<a=m(f,g),)$$ $$ text {and } quad g(c)=h(c)quad (text { as } c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.
$endgroup$
Suppose $f ne g ne h ne f.$ Let $a=m(f,g),,b=m(g,h), ,c=m(f,h).$
Suppose $d(f,g)+d(g,h)<d(f,h).$
Then $ 2^{-a}+2^{-b}<2^{-c},$ which implies $(,c<a land c<b,). $ But then $$ f(c)=g(c)quad (text { as }c<a=m(f,g),)$$ $$ text {and } quad g(c)=h(c)quad (text { as } c<b=m(g,h),)$$ so $f(c)=h(c),$ contrary to the def'n of $c=m(f,h).$ Contradiction.
edited 1 hour ago
answered 1 hour ago
DanielWainfleetDanielWainfleet
35.5k31648
35.5k31648
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
add a comment |
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
Thanks! So then perhaps the second part can be done by contradiction also?
$endgroup$
– Alexander Quinn
1 hour ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
$begingroup$
This is a very nice answer.
$endgroup$
– H. H.
24 mins ago
add a comment |
$begingroup$
If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min {m(a, b), m(b, c)} - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
$$m(a, c) ge min{m(a, b), m(b, c)}.$$
Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
begin{align*}
d(a, b) + d(b, c) &= 2^{-m(a, b)} + 2^{-m(b, c)} \
&= 2^{-max{m(a, b), m(b, c)}} + 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-m(a, c)} = d(a, c).
end{align*}
If $g in B(f, 2^{-n})$, then either $f = g$ (in which case, the result is trivially true), or
$$2^{-n} > d(f, g) = 2^{-m(a, b)} implies m(a, b) > n.$$
So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.
Note that the distance function $d(f, g)$ only takes countably many values:
$$0, frac{1}{2}, frac{1}{4}, ldots, frac{1}{2^n}, ldots$$
When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
$$Bleft(f, frac{1}{3}right) = Bleft(f, frac{1}{2}right) = B'left(f, frac{1}{4}right),$$
simply because a point that at most $frac{1}{3}$ distance away from $f$, basically only rules out their distances being $frac{1}{2}$.
This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.
$endgroup$
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
|
show 2 more comments
$begingroup$
If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min {m(a, b), m(b, c)} - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
$$m(a, c) ge min{m(a, b), m(b, c)}.$$
Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
begin{align*}
d(a, b) + d(b, c) &= 2^{-m(a, b)} + 2^{-m(b, c)} \
&= 2^{-max{m(a, b), m(b, c)}} + 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-m(a, c)} = d(a, c).
end{align*}
If $g in B(f, 2^{-n})$, then either $f = g$ (in which case, the result is trivially true), or
$$2^{-n} > d(f, g) = 2^{-m(a, b)} implies m(a, b) > n.$$
So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.
Note that the distance function $d(f, g)$ only takes countably many values:
$$0, frac{1}{2}, frac{1}{4}, ldots, frac{1}{2^n}, ldots$$
When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
$$Bleft(f, frac{1}{3}right) = Bleft(f, frac{1}{2}right) = B'left(f, frac{1}{4}right),$$
simply because a point that at most $frac{1}{3}$ distance away from $f$, basically only rules out their distances being $frac{1}{2}$.
This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.
$endgroup$
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
|
show 2 more comments
$begingroup$
If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min {m(a, b), m(b, c)} - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
$$m(a, c) ge min{m(a, b), m(b, c)}.$$
Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
begin{align*}
d(a, b) + d(b, c) &= 2^{-m(a, b)} + 2^{-m(b, c)} \
&= 2^{-max{m(a, b), m(b, c)}} + 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-m(a, c)} = d(a, c).
end{align*}
If $g in B(f, 2^{-n})$, then either $f = g$ (in which case, the result is trivially true), or
$$2^{-n} > d(f, g) = 2^{-m(a, b)} implies m(a, b) > n.$$
So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.
Note that the distance function $d(f, g)$ only takes countably many values:
$$0, frac{1}{2}, frac{1}{4}, ldots, frac{1}{2^n}, ldots$$
When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
$$Bleft(f, frac{1}{3}right) = Bleft(f, frac{1}{2}right) = B'left(f, frac{1}{4}right),$$
simply because a point that at most $frac{1}{3}$ distance away from $f$, basically only rules out their distances being $frac{1}{2}$.
This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.
$endgroup$
If $a(i)$ agrees with $b(i)$ for all $i = 1, ldots, m(a, b) - 1$. Similarly, $b(i)$ agrees with $c(i)$ for all $i = 1, ldots, m(b, c) - 1$. So, we have $a(i) = b(i) = c(i)$ for all $i = 1, ldots, k$, where $k = min {m(a, b), m(b, c)} - 1$, as $k$ is a number smaller than both $m(a, b) - 1$ and $m(b, c) - 1$. Therefore, $m(a, c)$, the first $i$ such that $a(i) neq c(i)$, must be at least $k + 1$. That is,
$$m(a, c) ge min{m(a, b), m(b, c)}.$$
Thus, excluding the easy cases where $a, b, c$ are not pairwise distinct,
begin{align*}
d(a, b) + d(b, c) &= 2^{-m(a, b)} + 2^{-m(b, c)} \
&= 2^{-max{m(a, b), m(b, c)}} + 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-min{m(a, b), (b, c)}} \
&ge 2^{-m(a, c)} = d(a, c).
end{align*}
If $g in B(f, 2^{-n})$, then either $f = g$ (in which case, the result is trivially true), or
$$2^{-n} > d(f, g) = 2^{-m(a, b)} implies m(a, b) > n.$$
So, the first point at which $f$ and $g$ disagree occurs past $n$, i.e. $f(i) = g(i)$ for $i = 1, ldots, n$.
Note that the distance function $d(f, g)$ only takes countably many values:
$$0, frac{1}{2}, frac{1}{4}, ldots, frac{1}{2^n}, ldots$$
When $r$ is between these values, the ball $B(f; r)$ will not grow or shrink when you vary $r$ by a small amount. For example, for any $f in X$,
$$Bleft(f, frac{1}{3}right) = Bleft(f, frac{1}{2}right) = B'left(f, frac{1}{4}right),$$
simply because a point that at most $frac{1}{3}$ distance away from $f$, basically only rules out their distances being $frac{1}{2}$.
This might give you an idea about how to go about attacking this third problem. I'll let you think about it. Let me know via comment if you need further help.
answered 1 hour ago
Theo BenditTheo Bendit
19.3k12353
19.3k12353
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
|
show 2 more comments
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
So the main thing in the last question is that we need to prove for any f ∈ X and radius r, the closure of the open ball of radius r around x is the closed ball of radius r.
$endgroup$
– Alexander Quinn
45 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
@AlexanderQuinn Well actually, every open ball is closed, and is equal to a closed ball! But. not necessarily the closed ball of the same radius as the open ball. The closure of $B(f, 1/2)$ is $B(f, 1/2) = B'(f, 1/4)$, not $B(f, 1/2)$ (which is equal to $X$).
$endgroup$
– Theo Bendit
31 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
So it must be the B(f,r) = B'(f,r) is the only reasonable one out of the two as if you take the limit you eventually approach 0 for r
$endgroup$
– Alexander Quinn
21 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
(In my previous comment, the third (and last) instance of $B(f, 1/2)$ should be $B'(f, 1/2)$)
$endgroup$
– Theo Bendit
19 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
$begingroup$
@AlexanderQuinn I'm not certain what you mean here. You will have $B(f, r) = B'(f, r)$ when $r$ is not $0$ or a negative power of $2$, as in this case, the difference between the sets (which is the sphere $S(f, r)$) is empty! If $r$ is not one of these numbers, then there are no points in the space of distance exactly $r$ from each other.
$endgroup$
– Theo Bendit
17 mins ago
|
show 2 more comments
$begingroup$
For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.
For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.
Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball
$endgroup$
add a comment |
$begingroup$
For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.
For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.
Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball
$endgroup$
add a comment |
$begingroup$
For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.
For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.
Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball
$endgroup$
For any $ain X$ and radius $r$, the closure of the open ball of radius $r$ around $a$ is the closed ball of radius $r$.
For any two distinct points $a,b$ in the space and any positive $epsilon$, there is a point $c$ within $epsilon$ of $b$, and closer to $a$ than $b$ is.
That is, for every $aneq b$ and $epsilongt 0$, there is $c$ with $d(c,b)<epsilon$ and $d(a,c)<d(a,b)$.
Proof. If the closed ball property holds, then fix any $a,b$ with $r=d(a,b)$. Since the closure of $B$ includes $b$, the second property follows. Conversely, if the second property holds, then if $r=d(a,b)$, then the property ensures that $b$ is in the closure of $B$, and so the closure of the open ball includes the closed ball
answered 38 mins ago
Alexander QuinnAlexander Quinn
265
265
add a comment |
add a comment |
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6
$begingroup$
Please do mathJax, so difficult to read your question.. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Yujie Zha
3 hours ago
1
$begingroup$
I was going to edit your question to format it properly, but then I noticed that you've been here for almost a year now (happy anniversary, in $5$ days time!) and have asked four prior questions that others have all formatted for you. So, unfortunately, this time I'm just going to down-vote. If you do format your question, give me a ping in the comments (@TheoBendit) and I'll remove the down-vote, attempt to answer the question, and give you an up-vote to boot. If you have any questions about using MathJax, ask Meta. Good luck!
$endgroup$
– Theo Bendit
2 hours ago
1
$begingroup$
Alexander, it really does not take much effort, but you have not put it in for 360 days! If you don't know how to do it, then learn it, or keep the MathJax documentation page open while you're typing! I did this for six months , and eventually rote learned all the simple commands. Take the time to do it, because your question ticks all the boxes and deserves attention and upvotes, except for its legibility. (Yes, you are starting to change it, that is good!)
$endgroup$
– астон вілла олоф мэллбэрг
2 hours ago
2
$begingroup$
@AlexanderQuinn Removing the down-vote for effort, but I'd like to see a little bit more done before I come in and clean it up for you. You can press "edit" on your previous questions to see how others have formatted them, and you can even try changing things to test how it affects the layout. Here are some simple commands too:
neq
= "$neq$",in
= "$in$",ge
= "$ge$". Try putting$
signs around most of the maths, and it will already look much better. If you try another edit, I'll finish it up for you if you wish.$endgroup$
– Theo Bendit
2 hours ago
2
$begingroup$
$What is B'???$
$endgroup$
– William Elliot
1 hour ago