# If you can prove this, you are a genius.

1. Oct 6, 2011

### Jamin2112

1. The problem statement, all variables and given/known data

Are you up to the challenge?

Let (X,d) be a metric space and let A be a nonempty subset of X. Define f:X→ℝ by f(x)=inf{d(x,a): a in A}. Prove that f is continuous.

2. Relevant equations

Definition. Let (X,d) and (Y,p) be metric spaces. A function f:X→Y is continuous at a point a in X provided that for each positive number ε there is a positive number δ such that if x is in X and d(a,x)<δ, then p(f(a),f(x))<ε. A function f:X→Y is continuous provided it is continuous at each point of X.

3. The attempt at a solution

All I've been able to do so far is understand the function f given in the problem. For example, we could have X = {1, 2, 3}, A = {1}, and d(x,a) = |x-a|; then f(1) = 0, f(2) = 1, and f(3) = 2. BUT THIS STEP FUNCTION ISN'T CONTINUOUS! So do I have to assume f is continuous?

2. Oct 6, 2011

### HallsofIvy

Your error is that you have completely ignored the topology, or metric, on X. Since X has only three members, any appropriate metric you assign will give a "discrete topology"- every subset is open. In this particular case, for any $0< \delta< 1$, $d(p,q)<\delta$ implies p= q. In particular, for any $0< \delta< 1$, $d(x, 1)< \delta$ implies x= 1 so that $f(x)= f(1)= 1$ and then $d(f(x), 1)= d(1,1)= 0< \epsilon$.

Yes, this function is continuous. That doesn't require a genius, it only requires knowing the definitions.

3. Oct 6, 2011

### micromass

Start by showing the inequality

$$\inf_{a\in A}{d(x,a)}\leq d(x,y)+\inf_{a\in A}{d(y,a)}$$

4. Oct 6, 2011

### DiracRules

I hope I'm not wrong but:

the definition does not say that f(x)=d(x,a), it says that $f(x)=inf_{a\in A}d(x,a)$.

So, in your example, f(1)=0 (since 1-1=0) f(2)=0 (since d(x,a) can be either 0 or 1, and the lowest is 1), and f(3)=0 (same reason)...

since $x\in X$ and $a\in A\subset X\Rightarrow a\in X, \forall x,\,\,inf_{a\in A}d(x,a)=0$

Is that correct?
If so, the proof is straightforward :D

5. Oct 6, 2011

### Jamin2112

Thanks for the responses so far! I'll chew on 'em and then come back when I have another question!

6. Oct 7, 2011

### Jamin2112

I see ......

So choose x in X, y in A in X, and ε > 0 such that ε > d(x,y).

Since d(a,x) ≤ d(a,y) + d(x,y), we know infa in Ad(a,x) ≤ infa in Ad(a,y) + d(x,y), meaning f(y) ≤ f(x) + d(x,y), or equivalently, f(y) - f(x) ≤ d(x,y).

Thus ε > d(x,y) forces f(y) - f(x) < ε.

(This seems wrong.)