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, Aqua Jul 5 at 5:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Cameron Buie, Hagen von Eitzen, Theo Bendit, postmortes, Aqua
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Welcome to MSE. Please add your own ideas and efforts concerning the topic in order to address more people and don't get downvotes. $\endgroup$ – Jan Jul 4 at 19:47
  • 1
    $\begingroup$ Hint: Is $\mathbb{R}$ a Hausdorff space? $\endgroup$ – guy3141 Jul 4 at 20:04

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.

  • $\begingroup$ 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... $\endgroup$ – Dog_69 Jul 4 at 23:20
  • $\begingroup$ ... 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. $\endgroup$ – 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 $X$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.