# A continuous function in Hausdorff space

1. Jan 23, 2009

### Symmetryholic

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

Let $$A \subset X$$; let $$f : A \rightarrow Y$$ be continuous; let Y be Hausdorff. Show that if f may be extended to a continuous function $$g: \overline{A} \rightarrow Y$$, then g is uniquely determined by f.

2. Relevant equations

3. The attempt at a solution

If f is a homeomorphism, we can say A is a Hausdorff space.
Just given a continuous function f, I am clueless how to start this problem.

2. Jan 24, 2009

### Unco

The first thing to do is to make sure you know the definitions of everything that is mentioned: continuity, Hausdorff (from your other post, you can tick this), extension, closure, uniquely determined.

The last one might not be in your topology text. It means that if we had another continuous extension $$h:\overline{A}\to Y$$ of f, then we must have g=h. So in that sense g is unique.

So your task is to suppose there are two such functions $$g:\overline{A}\to Y$$, $$h: \overline{A}\to Y$$, and to show that g(x) = h(x) for all x in $$\overline{A}\backslash A$$ (since we know they agree on A already).

The Hausdorffness of Y hints as to where to start in such a proof. Suppose that $$g(x)\neq h(x)$$ for some $$x\in\overline{A}\backslash A$$. Now you want to apply the given information to obtain a contradiction.

Remember to show us your work if you get stuck!

3. Jan 24, 2009

### Symmetryholic

My attempt to this problem is as follows:

Suppose there exist two continuous functions $$g,h:\overline{A} \rightarrow Y$$ that extend f at $$\overline{A}$$. Since $$g(x)=h(x)=f(x)$$ at $$x \in A$$ by definition, we shall show that $$g(x)=h(x)$$ at $$x \in \overline{A} \setminus A$$.
Assume $$g(x) \neq h(x)$$ at $$x \in \overline{A} \setminus A$$. Since Y is a Hausdorff space, we have two disjoint open sets U and V containing $$g(x)$$ and $$h(x)$$ at $$x \in A'$$ (A' denotes a derived set of A), respectively.
Since g and h are continuous functions, we have open sets $$g^{-1}(U)$$ and $$h^{-1}(V)$$ containing x at $$x \in A'$$.
We claim that $$O = g^{-1}(U) \cap h^{-1}(V) \cap A$$ is not empty. Since $$g^{-1}(U) \cap h^{-1}(V)$$ are open sets containing $$x \in A'$$ and any open set containing $$x \in A'$$ should intersect at A by definition of a limit point.
Let $$x' \in O$$.Then $$g(x') \neq h(x')$$ since $$x'$$ belongs to both $$g^{-1}(U)$$ and $$h^{-1}(V)$$ by assumption. Since x' also belongs to A, we have $$g(x') = h(x')$$, contradiction.
Thus, $$g(x) = h(x)$$ at $$x \in \overline{A} \setminus A$$, and we conclude that a continous function $$g: \overline{A} \rightarrow Y$$ is a unique extension of f at $$\overline{A}$$.

Any advice will be highly appreciated.

4. Jan 25, 2009

### Unco

Very good! It is fun when things begin to flow so nicely.

5. Jan 25, 2009

### Symmetryholic

Thank you for your corrections. It helped me a lot to solve this problem starting your hints.