Show that if a collection of continuous functions on $X$ separates points, then $X$ must be Hausdorff

Question

I have to show the following:

Suppose that $X$ is a topological space for which there is a collection of continuous real-valued functions on $X$ that separates points in $X$. Show that $X$ must be Hausdorff.

Just as a reminder, $f$ is said to separate points whenever for any $u$, $v$ $\in X$, if $u \neq v$, then $f(u) \neq f(v)$, and a topological space $X$ is said to be Hausdorff provided that each two points in $X$ can be separated by disjoint neighborhoods.

I attempted to approach this problem by showing that Not Hausdorff $\mathbf{\to}$ Doesn't Separate Points. My attempt is as follows:

Suppose $X$ is not Hausdorff. Then, for some $u$, $v$ $\in X$ s.t. $u \neq v$ , $\displaystyle \exists N_{\delta_{1}}(u)$ and $\displaystyle N_{\delta_{2}}(v)$ s.t. $\displaystyle N_{\delta_{1}}(u) \cap N_{\delta_{1}}(v) \neq \emptyset$. I.e., $\displaystyle \exists x \in N_{\delta_{1}}(u) \cap N_{\delta_{2}}(v)$.

This implies that $|x-u|<\delta_{1}$, $|x-v|<\delta_{2}$. So, $|u-v| = |u-x+x-v| \leq |x-u| + |x-v| < \delta_{1}+\delta_{2} = \delta$. Then, because $f$ is continuous, this implies that $\forall \epsilon > 0$, $|f(u)-f(v)|<\epsilon$.

Since this is true for any $\epsilon > 0$, we conclude that $|f(u)-f(v)|= 0$, which implies that $f(u) = f(v)$, so $f$ does not separate points.

Is this proof correct, and if not, what is wrong with it? How can I fix it so that it is correct?

Answer 1

Suppose $\mathscr{F}$ is a collection of continuous functions $X\to \Bbb R$ that separates points, and suppose $x,y\in X, x\ne y$. By hypothesis, there is $f\in \mathscr{F}$ such that $f(x) \ne f(y)$. As $\Bbb R$ is Hausdorff, there are disjoint open neighborhoods $U, V$ of $f(x), f(y)$ respectively. Because $f$ is continuous, $f^{-1}(U)$ is open in $X$ and is a neighborhood of $x$, and similarly, $f^{-1}(V)$ is an open neighborhood of $y$. Clearly, $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint. So $X$ is Hausdorff.

Answer 2

Let $x,y\in X$, $f$ continuous such $f(x)\neq f(y)$. We suppose $f(x)<f(y)$, set $U=f^{-1}(-\infty,({{f(x)+f(y)}\over 2})$ and $V=f^{-1}({{f(x)+f(y)}\over 2},+\infty)$. $U$ and $V$ are open and $x\in U$, $x$ is not in $V$, $y\in V$ $y$ is not in $U$.

Answer 3

Direct proof would be much more easier. Here is how you can start: Let $f:X\to \mathbb R$ be a separating continuous function. Given $u\neq v\in X$, we have $f(u)\neq f(v)$ in $\mathbb R$. Since $\mathbb R$ is Hausdorff, can take disjoint neighborhoods $U,V$ of $f(u),f(v)$ in $\mathbb R$, respectively.

Now you try to complete the proof using $U$ and $V$.

As commented before, $X$ is a topological space, not necessarily a metric space. So, talking about the distance and taking $\delta$ balls doesn't make sense. Even when it makes sense, it makes the proof more complicated. When you examine a topological object, just take open sets.