Prove Continuous Function is Uniformly Continuous, Given Compact Subset Depending on Epsilon

I am working an old qual problem and am struggling with the following question:

Let $X, Y$ be metric spaces, and let $f : X \to Y$ be continuous. Suppose that for every $\epsilon > 0$, there exists a compact set $K_{\epsilon} \subset X$ such that $d(f(x), f(y)) < \epsilon$ whenever $x, y \in K_{\epsilon}^{C}$ (the complement of $K_{\epsilon}$). Show that $f$ is uniformly continuous.

Here is my attempt at a proof:

Proof: Let $\epsilon > 0$. Let $K_{\epsilon} \subset X$ be given, as in the assumptions. We will exhibit a $\delta > 0$ such that whenever $x, y \in X$ with $d(x, y) < \delta$, then $d(f(x), f(y)) < \epsilon$. We have the following cases:

Case 1: Suppose $x, y \in K_{\epsilon}^{C}$. Then by assumption, $d(f(x), f(y)) < \epsilon$. So $f$ is uniformly continuous on $K_{\epsilon}^{C}$.

Case 2: Suppose $x, y \in K_{\epsilon}$. As $K_{\epsilon}$ is compact and $f$ is continuous, we have that $f$ is uniformly continuous on $K_{\epsilon}$. So there exists $\delta_{K} > 0$ s.t. if $d(x, y) < \delta_{K}$, then $d(f(x), f(y)) < \epsilon$.

Case 3: Suppose $x \in K_{\epsilon}$ and $y \not \in K_{\epsilon}$. We note that the following is an open cover of $K_{\epsilon}$:

$$\bigcup_{x \in K_{\epsilon}} B_{\delta_{x}}(x) \, (\delta_{x} \in (0, \delta_{K}))$$

So by compactness, there exists $x_{1}, \ldots, x_{n}$ s.t.: $$K_{\epsilon} \subset \bigcup_{i=1}^{n} B_{\delta_{i}}(x_{i})$$

Now as $f$ is uniformly continuous on $K$, there exists $\delta_{0} > 0$ s.t. if $d(a, b) < \delta_{0}$ then $d(f(a), f(b)) < \frac{\epsilon}{n}$.

Fix a point $z \in \partial K_{\epsilon}$. As $f$ is continuous, there exists $\delta_{1} > 0$ s.t. if $d(w, z) < \delta_{1}$, then $d(f(w), f(z)) < \epsilon$. Fix such a $w \in (K_{\epsilon}^{C} \cap B_{\delta_{1}}(z))$.

As $w \in K_{\epsilon}^{C}$, $d(f(w), f(y)) < \epsilon$.

Thus: $d(f(x), f(y)) \leq d(f(x), f(z)) + d(f(z), f(w)) + d(f(w), f(y)) = 3\epsilon$.

After cleaning up case 3, it is simply a matter of massaging my deltas and epsilons, and picking the smallest delta. Where I am really stuck is how to control the distance (in $X$) between $x$ and $z$. If I can do that, then the rest of the proof should come together nicely. Does anybody have any suggestions or pointers on how to go about this? Alternatively, is there a better way of doing this?

Thank you in advance for any help!

• My proof for Case 3 is incomplete. I desire the control on $d(f(x), f(z))$, but I still need to justify why this control is valid. Since $x, z \in K_{\epsilon}$, I want to use the uniform continuity of $f$ on $K_{\epsilon}$. How do I show that $d(x, z) < \delta_{K}$? Or is there a better way of justifying that $d(f(x), f(z)) < \epsilon$? – ml0105 Nov 24 '17 at 1:40

1 Answer

Here's an alternative proof for your third case:

Assume that there exists $(x_n)$ and $(y_n)$ such that $x_n \in K_\epsilon$, $y_n \in K_\epsilon^c$ such that $d(x_n, y_n) < 1/n$ but $d(f(x_n), f(y_n)) \geq \epsilon$. Compactness of $K_\epsilon$ implies the existence of a convergent subsequence of $(x_n)$, so we can assume WLOG that $(x_n)$ converges to some $x$. Then,

$$d(x, y_n) \leq d(x, x_n) + d(x_n, y_n) < d(x, x_n) + \frac{1}{n}$$

by the triangle inequality. We can make $d(x, x_n)$ arbitrarily small, so $y_n$ also converges to $x$.

Now, $f$ is continuous at $x$, so there exists some $\delta > 0$ such that $d(p, x) < \delta$ implies that $d(f(p), f(x)) < \epsilon / 2$. Since both $(x_n)$ and $(y_n)$ converge to $x$, there will be a pair $x_k, y_k$ such that both are contained in the ball of radius $\delta$ centered at $x$. However, $d(f(x_k), f(y_k)) \geq \epsilon$. Contradiction.

• I really like this sequential argument! Thanks for sharing it! – ml0105 Jan 2 '18 at 17:41