# If $\mathcal C(X,\Bbb R)$ separates the points of $X$, then prove that $X$ is Hausdorff. [closed]

Let $$X$$ be a topological space and consider the set $$\mathcal C(X,\Bbb R)$$ which is the set of all bounded continuous real valued functions on $$X$$.

I am just starting with the subject and would thus like to see a rigorous proof of this so as to get acquainted with the proof methodology.

## closed as off-topic by Cameron Buie, Hagen von Eitzen, Theo Bendit, postmortes, AquaJul 5 at 5:44

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• Welcome to MSE. Please add your own ideas and efforts concerning the topic in order to address more people and don't get downvotes. – Jan Jul 4 at 19:47
• Hint: Is $\mathbb{R}$ a Hausdorff space? – guy3141 Jul 4 at 20:04

## 2 Answers

First; what does it mean for a family of functions to separate points?

If $$\mathscr{F}$$ is a family of functions defined on some set or space X, then $$\mathscr{F}$$ is said to separate points if $$\forall x,y\in X$$ with $$x\neq y$$, $$\exists f\in \mathscr{F}$$ such that $$f(x)\neq f(y)$$

In your problem $$\mathscr{F}=\mathscr{C}(X,\mathbb{R})$$

To prove this we also need to recall the definition of a Hausdorff space.

X is Hausdorff if $$\forall x,y \in X$$ distinct $$\exists U,V$$ open (or in the topology of X if you like) so that $$x\in U, y\in V$$ and $$U \cap V =\emptyset$$

Now that we have the definitions down the proof is fairly straight forward.

Suppose $$x,y \in X$$ distinct ($$x\neq y$$) Since $$\mathscr{C}(X,\mathbb{R})$$ separates points we can find $$f \in \mathscr{C}(X,\mathbb{R})$$ so that $$f(x) \neq f(y)$$

We know that $$\mathbb{R}$$ is a Hausdorff Space (try and prove it yourself). So we can find $$U , V$$ open and disjoint with $$f(x) \in U, f(y) \in V$$ Since f is a continuous function from X to $$\mathbb{R}$$ and $$U,V$$ open we know $$f^{-1}(U)$$ and $$f^{-1}(V)$$ are open.
Try to verify that $$f^{-1}(U)$$ and $$f^{-1}(V)$$ are disjoint. If the are disjoint then we are done since we have found disjoint open sets each containing a distinct point in X. Since the pair of points were arbitrary we are done and can say that X is a Hausdorff space.

A few take aways from this problem.

-The continuous functions on a topological space define the topology and vice versa. What does this mean? When you give me a topology, you tell me which functions are continuous. When you give me the family of continuous functions you tell me which sets are open and hence give me a topology.

Often times in functional analysis we will choose a topology to work with by declaring a certain family of functions are continuous.

-The assumption that the family separates points was essential in concluding that X is Hausdorff.

-Finally, if you are studying topology I highly recommend reading about the separation axioms and Urysohn Lemma.

Hope this helped, take care.

• Certainly it is a fantastic answer. However, I would like to remark some aspect I think may be misunderstood here. I hope not to be so pedantic. Even when topology determines who the continues functions are, it is important to highlight that the set $C(X,\mathbb R)$ may not be unique (in fact it isn't). Given Amy topological space $X$, there exists another one, $Y$, such that $C(X,\mathbb R)$ and $C(Y,\mathbb R)$ are isomorphic and moreover $Y$ is a Tychonoff space. I learnt that result by eading the Willard, book which I recommend you... – Dog_69 Jul 4 at 23:20
• ... if you are interested in topology @AbhigyanSaha. However, as Henno told me once, the main reference for this sort of results is Gillman and Jerison "Rings of continuous functions", altough I don't like the proof they give. I prefer my reasoning which I asked about here and here. – Dog_69 Jul 4 at 23:25

Let $$x_{1}$$ and $$x_{2}$$ be two distinct elements of $$X$$. By hypothesis, there exists a continuous function $$f \colon X \rightarrow \mathbb{R}$$ such that $$f\left( x_{1} \right) \neq f\left( x_{2} \right)$$. As $$\mathbb{R}$$ is Hausdorff -- it is a metric space -- there exists a neighbourhood $$V_{1}$$ of $$f\left( x_{1} \right)$$ in $$\mathbb{R}$$ and a neighbourhood $$V_{2}$$ of $$f\left( x_{2} \right)$$ in $$\mathbb{R}$$ such that $$V_{1} \cap V_{2} = \varnothing$$. Now, set $$U_{1} = f^{-1}\left( V_{1} \right)$$ and $$U_{2} = f^{-1}\left( V_{2} \right)$$. We have $$U_{1} \cap U_{2} = \varnothing$$, and, since $$f$$ is continuous, $$U_{1}$$ is a neighbourhood of $$x_{1}$$ in $$X$$ and $$U_{2}$$ is a neighbourhood of $$x_{2}$$ in $$X$$.