Hint for this equicontinuity/bounded question
up vote
1
down vote
favorite
Let $X$ be a compact metric space. Prove that an equicontinuous subset
of $C(X)$ is uniformly bounded if it is pointwise bounded.
I'm not looking for the answer. I know all the relevant definitions but I'd like a hint as to what $X$ being compact implies.
real-analysis general-topology
edited Nov 13 at 17:32
Bernard
115k637108
asked Nov 13 at 17:23
Thomas
682415
add a comment |
up vote
1
down vote
favorite
Let $X$ be a compact metric space. Prove that an equicontinuous subset
of $C(X)$ is uniformly bounded if it is pointwise bounded.
I'm not looking for the answer. I know all the relevant definitions but I'd like a hint as to what $X$ being compact implies.
real-analysis general-topology
edited Nov 13 at 17:32
Bernard
115k637108
asked Nov 13 at 17:23
Thomas
682415
As a way to understand why you need compactness, can you create a counterexample if compactness is lacking?
– Ted Shifrin
Nov 13 at 17:49
On $mathbb{R}$, the one-member family consisting of $f(x)=x$ works. Or the family consisting of $f(x)=ax$ for all $|a|leq 100$.
– user25959
Nov 13 at 18:08
@user25959: I was addressing that to the original questioner, not to the general public.
– Ted Shifrin
Nov 13 at 18:09
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a compact metric space. Prove that an equicontinuous subset
of $C(X)$ is uniformly bounded if it is pointwise bounded.
I'm not looking for the answer. I know all the relevant definitions but I'd like a hint as to what $X$ being compact implies.
real-analysis general-topology
edited Nov 13 at 17:32
Bernard
115k637108
asked Nov 13 at 17:23
Thomas
682415
Let $X$ be a compact metric space. Prove that an equicontinuous subset
of $C(X)$ is uniformly bounded if it is pointwise bounded.
I'm not looking for the answer. I know all the relevant definitions but I'd like a hint as to what $X$ being compact implies.
real-analysis general-topology
real-analysis general-topology
edited Nov 13 at 17:32
Bernard
115k637108
asked Nov 13 at 17:23
Thomas
682415
edited Nov 13 at 17:32
Bernard
115k637108
asked Nov 13 at 17:23
Thomas
682415
edited Nov 13 at 17:32
Bernard
115k637108
edited Nov 13 at 17:32
Bernard
115k637108
edited Nov 13 at 17:32
Bernard
115k637108
115k637108
asked Nov 13 at 17:23
Thomas
682415
asked Nov 13 at 17:23
Thomas
682415
asked Nov 13 at 17:23
Thomas
682415
682415
As a way to understand why you need compactness, can you create a counterexample if compactness is lacking?
– Ted Shifrin
Nov 13 at 17:49
On $mathbb{R}$, the one-member family consisting of $f(x)=x$ works. Or the family consisting of $f(x)=ax$ for all $|a|leq 100$.
– user25959
Nov 13 at 18:08
@user25959: I was addressing that to the original questioner, not to the general public.
– Ted Shifrin
Nov 13 at 18:09
add a comment |
As a way to understand why you need compactness, can you create a counterexample if compactness is lacking?
– Ted Shifrin
Nov 13 at 17:49
On $mathbb{R}$, the one-member family consisting of $f(x)=x$ works. Or the family consisting of $f(x)=ax$ for all $|a|leq 100$.
– user25959
Nov 13 at 18:08
@user25959: I was addressing that to the original questioner, not to the general public.
– Ted Shifrin
Nov 13 at 18:09
As a way to understand why you need compactness, can you create a counterexample if compactness is lacking?
– Ted Shifrin
Nov 13 at 17:49
As a way to understand why you need compactness, can you create a counterexample if compactness is lacking?
– Ted Shifrin
Nov 13 at 17:49
On $mathbb{R}$, the one-member family consisting of $f(x)=x$ works. Or the family consisting of $f(x)=ax$ for all $|a|leq 100$.
– user25959
Nov 13 at 18:08
On $mathbb{R}$, the one-member family consisting of $f(x)=x$ works. Or the family consisting of $f(x)=ax$ for all $|a|leq 100$.
– user25959
Nov 13 at 18:08
@user25959: I was addressing that to the original questioner, not to the general public.
– Ted Shifrin
Nov 13 at 18:09
@user25959: I was addressing that to the original questioner, not to the general public.
– Ted Shifrin
Nov 13 at 18:09
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
We'll denote members of the equicontinuous family by $f_alpha$.
Let $M_x >0$ satisfy $|f_alpha(x)|< M_x$ for all $alpha$ (which exists because of the pointwise bound).
Now fix $x$ in $X$. Equicontinuity implies there is some $delta$ such that $d(x,y)< delta implies |f_alpha(y)| < M_{x} + 1$.
Consider the set $B_x = B(x,delta)$. It is (and here's the hint) an open set containing $x$.
answered Nov 13 at 17:37
user25959
1,016714
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
1
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
1
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
1
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
|
show 7 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We'll denote members of the equicontinuous family by $f_alpha$.
Let $M_x >0$ satisfy $|f_alpha(x)|< M_x$ for all $alpha$ (which exists because of the pointwise bound).
Now fix $x$ in $X$. Equicontinuity implies there is some $delta$ such that $d(x,y)< delta implies |f_alpha(y)| < M_{x} + 1$.
Consider the set $B_x = B(x,delta)$. It is (and here's the hint) an open set containing $x$.
answered Nov 13 at 17:37
user25959
1,016714
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
1
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
1
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
1
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
|
show 7 more comments
up vote
1
down vote
We'll denote members of the equicontinuous family by $f_alpha$.
Let $M_x >0$ satisfy $|f_alpha(x)|< M_x$ for all $alpha$ (which exists because of the pointwise bound).
Now fix $x$ in $X$. Equicontinuity implies there is some $delta$ such that $d(x,y)< delta implies |f_alpha(y)| < M_{x} + 1$.
Consider the set $B_x = B(x,delta)$. It is (and here's the hint) an open set containing $x$.
answered Nov 13 at 17:37
user25959
1,016714
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
1
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
1
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
1
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
|
show 7 more comments
up vote
1
down vote
up vote
1
down vote
We'll denote members of the equicontinuous family by $f_alpha$.
Let $M_x >0$ satisfy $|f_alpha(x)|< M_x$ for all $alpha$ (which exists because of the pointwise bound).
Now fix $x$ in $X$. Equicontinuity implies there is some $delta$ such that $d(x,y)< delta implies |f_alpha(y)| < M_{x} + 1$.
Consider the set $B_x = B(x,delta)$. It is (and here's the hint) an open set containing $x$.
answered Nov 13 at 17:37
user25959
1,016714
We'll denote members of the equicontinuous family by $f_alpha$.
Let $M_x >0$ satisfy $|f_alpha(x)|< M_x$ for all $alpha$ (which exists because of the pointwise bound).
Now fix $x$ in $X$. Equicontinuity implies there is some $delta$ such that $d(x,y)< delta implies |f_alpha(y)| < M_{x} + 1$.
Consider the set $B_x = B(x,delta)$. It is (and here's the hint) an open set containing $x$.
answered Nov 13 at 17:37
user25959
1,016714
edited Nov 13 at 17:41
answered Nov 13 at 17:37
user25959
1,016714
answered Nov 13 at 17:37
user25959
1,016714
answered Nov 13 at 17:37
user25959
1,016714
1,016714
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
1
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
1
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
1
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
|
show 7 more comments
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
1
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
1
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
1
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Why is $B_x$ an open set? Maybe better to let $B_x=B(x,delta)$?
– Anguepa
Nov 13 at 17:41
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
Thank you for the comment. It is open but that would require another equicontinuity argument. I changed it.
– user25959
Nov 13 at 17:42
1
1
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
Definition of compactness: cover $X$ with the $B_x$. Then extract a finite subcover.
– user25959
Nov 13 at 17:49
1
1
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
That's what I was thinking @Thomas. Even though $X$ is infinite, every $y$ can be placed into some $B(x_i,delta_{x_i})$ and hence by the construction of $delta_{x_i}$, etc... If $X$ were finite then the uniform bound would come out almost immediately (can you think why). Compactness when $X$ is infinite makes the argument almost the same argument as in the finite case.
– user25959
Nov 13 at 18:22
1
1
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
Notice how we constructed $B_x$ using an arbitrary $x$. Then the open cover is ${B_x}$ for all x.
– user25959
Nov 13 at 19:26
|
show 7 more comments
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997020%2fhint-for-this-equicontinuity-bounded-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
As a way to understand why you need compactness, can you create a counterexample if compactness is lacking?
– Ted Shifrin
Nov 13 at 17:49
On $mathbb{R}$, the one-member family consisting of $f(x)=x$ works. Or the family consisting of $f(x)=ax$ for all $|a|leq 100$.
– user25959
Nov 13 at 18:08
@user25959: I was addressing that to the original questioner, not to the general public.
– Ted Shifrin
Nov 13 at 18:09