# A continuous function in Hausdorff space

• Symmetryholic
In summary, if you have a continuous function and you want to extend it to a new domain, you need to find a way to uniquely determine the extension.
Symmetryholic

## Homework Statement

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.

## 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.

Symmetryholic said:

## Homework Statement

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.

## 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.
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!

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 continuous function $$g: \overline{A} \rightarrow Y$$ is a unique extension of f at $$\overline{A}$$.

Any advice will be highly appreciated.

Symmetryholic said:

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 $$g(x) \neq h(x)$$ (for some) $$x \in \overline{A} \setminus A$$. 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, $$g(x) = h(x)$$ (for all) $$x \in \overline{A} \setminus A$$, and we conclude that a continuous function $$g: \overline{A} \rightarrow Y$$ is a unique extension of f at $$\overline{A}$$.
Very good! It is fun when things begin to flow so nicely.

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

## 1. What is a Hausdorff space?

A Hausdorff space is a type of topological space in which any two distinct points have disjoint neighborhoods. In other words, there is always a way to separate two points in a Hausdorff space using open sets.

## 2. What is a continuous function?

A continuous function is a function between two topological spaces that preserves the topological structure. This means that if two points are close together in the domain, their images will also be close together in the range. In other words, small changes in the domain result in small changes in the range.

## 3. How is continuity defined in a Hausdorff space?

A function is continuous in a Hausdorff space if and only if the preimage of any closed set is also closed. In simpler terms, this means that if a set is closed in the range, its inverse image will also be closed in the domain.

## 4. What are some examples of continuous functions in Hausdorff spaces?

Examples of continuous functions in Hausdorff spaces include polynomial functions, trigonometric functions, and exponential functions. Any function that can be drawn without lifting your pencil from the paper is also continuous in a Hausdorff space.

## 5. Why is the concept of continuous functions in Hausdorff spaces important?

Continuous functions in Hausdorff spaces are important because they provide a way to understand and analyze the behavior of functions in a topological space. They allow us to make predictions about the behavior of a function based on its properties, and they are essential in many areas of mathematics, including analysis, topology, and geometry.

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