Showing the continuity of $d(x,f(x))$

Question

Assume that $(X,d)$ is compact, and that $f: X \to X$ is continuous. Show that the function $g(x) = d(x,f(x))$ is continuous and has a minimum point.

Consider the function $g(x) = d(x,f(x))$. If $g$ is continuous, then $\forall \epsilon >0 \ \exists \delta > 0$ such that $$d(x,y) < \delta \implies d(g(x), g(y)) < \epsilon.$$ Since $f$ is continuous and that $X$ is compact, we have that $f: X \to X$ is uniformly continuous. Therefore, $\forall \epsilon > 0 \ \exists \delta > 0 \ \text{such that} \ d(x,y) < \delta \implies d(f(x),f(y))< \epsilon.$

But how do we go on from here, considering $d(g(x),g(y)) = d(d(x,f(x)),d(y,f(y)))$ gets quite messy?

Notice that $g$ is a real valued function and the standard metric in $\mathbb{R}$ is $|\cdot-\cdot|$ — clark, Jun 13 '16 at 06:04
@Jordan: After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. Thank you :-) — Watson, Jun 18 '16 at 18:14

score 2 · Answer 1 · edited Apr 13 '17 at 12:19

You could also do this in steps:

First you could show that the map $$(x,x)\mapsto (x,f(x))$$ is a continuous function from $X\times X$ to $X\times X$.
Then you could show that $$(x,y)\mapsto d(x,y)$$ is a continuous function from $X\times X$ to $\mathbb R$.
Your function $g$ is the composition of the above two functions. Composition of continuous function is continuous.

Remark: If this is your first encounter with metric spaces, then you are probably expected to use $\epsilon$-$\delta$ definition and work directly with the map $g(x)$. Later, as you learn more about topological space (in particular, when you learn about products spaces) this approach will seem more natural to you.

The claim from the first bullet point can be generalized: Show that if $f,g$ are continuous then so is $f\times g$.

The claim from the second bullet point was discussed in several posts here on math.SE. For example, Metric is continuous function.

score 1 · Answer 2 · edited Jun 13 '16 at 07:16

As noted in the comments, $g: X \rightarrow \mathbb R$. Drawing a picture can be helpful. Since $f$ is continuous, when $x$ and $y$ are close to eachother $f(x)$ and $f(y)$ are close to each other, so $d(x, f(x))$ and $d(y,f(y))$ cannot differ by a large amount. We can formalize an argument as follows, using the triangle inequality, naturally:

$d(x,f(x)) \leq d(f(x), f(y)) + d(x,f(y)) \leq d(f(x), f(y)) + d(x,y) + d(y,f(y)).$

Similarly,

$d(y,f(y))) \leq d(f(x), f(y)) + d(x,y) + d(x,f(x)).$

It follows that

$|d(x,f(x)) - d(y,f(y))| \leq d(f(x), f(y)) + d(x,y).$

Now take $\min(\epsilon, \delta)$ as your $\delta$ (the $\delta$ comes from the uniform continuity of $f$). So $g$ is continuous. Since $X$ is compact, it attains a minimum.

Showing the continuity of $d(x,f(x))$

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