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Continuous mapping of compact metric spaces

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$ then $f$ is uniformly continuous on $X$.

I have seen a proof in the Rudin's book but I don't quite get it , can anybody establish another proof but with more details ?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492

TheBigBadBen

Active member
May 12, 2013
84
We could also prove this using the topological definition of compactness (i.e. that every open cover has a finite subcover) rather than sequential compactness (i.e. that every sequence has a convergent subsequence). To me, this proof is nicer, though I can't guarantee it will be any easier on the intuition.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
We could also prove this using the topological definition of compactness (i.e. that every open cover has a finite subcover) rather than sequential compactness (i.e. that every sequence has a convergent subsequence). To me, this proof is nicer, though I can't guarantee it will be any easier on the intuition.
But uniform continuity is not defined in a general topological space, so any proof will have to refer to the metric at some point.
 

TheBigBadBen

Active member
May 12, 2013
84
But uniform continuity is not defined in a general topological space, so any proof will have to refer to the metric at some point.
Right. Here's a sketch of the proof I have in mind:

Given a continuous $f:X\to Y$, we want to show that for any $\epsilon>0$, there is a $\delta>0$ so that $d_Y(f(x),f(y))<\epsilon$ whenever $d_X(x,y)<\delta$

Consider any $\epsilon>0$. By continuity, we may state that for each $x\in X$, there is a $\delta_x$ such that for any $y \in X$, $d_Y(f(x),f(y))<\epsilon/2$ when $d_X(x,y)<\delta_x$. Now, consider the open cover given by
$$\left\{ B_{\delta_x}(x)| x\in X\right\}$$
(where $B_{r}(x)$ the open ball of radius $r$ and center $x$).

By compactness, there exists a finite subcover of the form

\[\left\{ B_{\delta_{x_k}}(x_k)|k\in\{1,2,\dots,n\}\right\}\]

Since there are finitely many $x_k$, there is a minimum $\delta_{x_k}$, which we may call $\delta$. We may now state (via some triangle-inequality magic) that $d_Y(f(x),f(y))<\epsilon$ whenever $d_X(x,y)<\delta$.

Thus, we have shown $f$ to be uniformly continuous
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Right. Here's a sketch of the proof I have in mind:

Given a continuous $f:X\to Y$, we want to show that for any $\epsilon>0$, there is a $\delta>0$ so that $d_Y(f(x),f(y))<\epsilon$ whenever $d_X(x,y)<\delta$

Consider any $\epsilon>0$. By continuity, we may state that for each $x\in X$, there is a $\delta_x$ such that for any $y \in X$, $d_Y(f(x),f(y))<\epsilon/2$ when $d_X(x,y)<\delta_x$. Now, consider the open cover given by
$$\bigcup_{x\in X} B_{\delta_x}(x)$$
(where $B_{r}(x)$ the open ball of radius $r$ and center $x$).

By compactness, there exists a finite subcover of the form
$$\bigcup_{k=1}^n B_{\delta_{x_k}}(x_k)$$
Since there are finitely many $x_k$, there is a minimum $\delta_{x_k}$, which we may call $\delta$. We may now state (via some triangle-inequality magic) that $d_Y(f(x),f(y))<\epsilon$ whenever $d_X(x,y)<\delta$.

Thus, we have shown $f$ to be uniformly continuous
The triangle inequality is a powerful tool, but its magic is a bit more subtle than that. In fact, those open balls $B_{\delta_{x_k}}(x_k)$ cover $X$. So if you are given $x,y\in X$ with $d_X(x,y)<\delta$, it follows that $x$ and $y$ must each lie in one of them, say $x\in B_{\delta_{x_i}}(x_i)$ and $y\in B_{\delta_{x_j}}(x_j)$. But there is no guarantee that $x$ and $y$ belong to the same ball (in other words, you can't assume that $i=j$). You cannot then deduce that $d_Y(f(x),f(y))<\epsilon$.

What you have to do is this. Given $\varepsilon>0$, define $B_{\delta_x}(x)$ as before, for each $x\in X$. Then consider the cover of $X$ consisting of balls of half that radius. The collection $\{B_{\delta_x/2}(x)\}$ has a finite subcover. Let $\delta$ be the minimum of the $\delta_{x_k}$s, as before. Then if $d_X(x,y)<\delta/2$ you can conclude that $x\in B_{\delta_{x_k/2}}(x_k)$ for some $k$. It follows from the triangle inequality that $x$ and $y$ are both in $B_{\delta_{x_k}}(x_k)$, from which you can conclude that $d_Y(f(x),f(y))\leqslant d_Y(f(x),f(x_k)) + d_Y(f(x_k),f(y)) < 2\varepsilon$. That is enough to establish uniform continuity.
 

TheBigBadBen

Active member
May 12, 2013
84
The triangle inequality is a powerful tool, but its magic is a bit more subtle than that. In fact, those open balls $B_{\delta_{x_k}}(x_k)$ cover $X$. So if you are given $x,y\in X$ with $d_X(x,y)<\delta$, it follows that $x$ and $y$ must each lie in one of them, say $x\in B_{\delta_{x_i}}(x_i)$ and $y\in B_{\delta_{x_j}}(x_j)$. But there is no guarantee that $x$ and $y$ belong to the same ball (in other words, you can't assume that $i=j$). You cannot then deduce that $d_Y(f(x),f(y))<\epsilon$.

What you have to do is this. Given $\varepsilon>0$, define $B_{\delta_x}(x)$ as before, for each $x\in X$. Then consider the cover of $X$ consisting of balls of half that radius. The collection $\{B_{\delta_x/2}(x)\}$ has a finite subcover. Let $\delta$ be the minimum of the $\delta_{x_k}$s, as before. Then if $d_X(x,y)<\delta/2$ you can conclude that $x\in B_{\delta_{x_k/2}}(x_k)$ for some $k$. It follows from the triangle inequality that $x$ and $y$ are both in $B_{\delta_{x_k}}(x_k)$, from which you can conclude that $d_Y(f(x),f(y))\leqslant d_Y(f(x),f(x_k)) + d_Y(f(x_k),f(y)) < 2\varepsilon$. That is enough to establish uniform continuity.
Ah, I knew something was off about my proof. Thank you for picking that up and wrapping it up neatly, and for imparting some of your own triangle-inequality magic.

At any rate, I prefer to think of compactness in this sense rather than in terms of convergent subsequence, and I think this proof has a certain directness that the others lack. That might just be me though.