Continuous mapping of compact metric spaces

[ZaidAlyafey]

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]

What about the Wikipedia proof and the link to PlanetMath therein?

 

[TheBigBadBen]

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]

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]

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

 

[Opalg]

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]

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.