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I came across with a theorem on Folland's Real Analysis. The theorem is as follows:

Let $X$ be a set, $F:$ collection of $\mathbb{R}$-valued functions on $X$, $\tau$: weak topology generated by $F$.(The smallest topology on $X$ contanining $F$). Then, $\tau$ is Hausdorff $\iff \forall x,y \in X $ with $x \neq y$, $\exists f \in F$ s.t. $f(x)\neq f(y)$.

I proved the $(\impliedby)$ part. I am taking distinct $x,y$ elements and I have that $f(x) \neq f(y)$. I take two disjoint open sets in $\mathbb{R}$ s.t. one contains $f(x)$ and the other contains $f(y)$. Then the preimages of these sets give us two disjoint open sets containing $x$ and $y$ relatively.

However, I could not prove the $(\implies)$ part and need some help.

Ninja
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  • How does a basic neighbourhood of $x$ look like? – Daniel Fischer Mar 01 '17 at 12:11
  • There are some points contained in an open set where $y$ does not belong to this set. – Ninja Mar 01 '17 at 12:18
  • How is the topology defined? From the collection of functions, how does one construct a basis of the topology? – Daniel Fischer Mar 01 '17 at 12:31
  • Perhaps the weak topology generated by $F$ should, instead, be the smallest topology on $X$ containing $f^{-1}(U)$ for each $f \in F$ and each open $U \subset \mathbb{R}$; equivalently, the smallest topology on $X$ such that each $f \in F$ is continuous. – Lee Mosher Mar 01 '17 at 12:41
  • Can we say that for some $f$, $B={f^{-1}(n,n+1) : n \in \mathbb{Z} }$ is a basis? – Ninja Mar 01 '17 at 13:01
  • @LeeMosher, how can we say it? – Ninja Mar 01 '17 at 13:03
  • No, it is not sufficient to pick out just one $f$ and just the open intervals $(n,n+1) \subset \mathbb{R}$ with integer endpoints. You must use all $f \in F$ and all open intervals $(a,b) \subset \mathbb{R}$ with all possible endpoints $a<b$. – Lee Mosher Mar 01 '17 at 13:06
  • I am very confused now, unfortunately. Now I take two distinct points $x,y \in X$ and two disjoint open sets $U,V$ containing $x$ and $y$ relatively. So our open sets are consisting of functions but now I am saying that the elements $x$ and $y$ belong to these sets. How can it be possible? – Ninja Mar 01 '17 at 13:18

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In this long answer I explain what a base for this topology generated by $\mathcal{F}$ looks like: sets of the form $f^{-1}[O]$ where $f \in \mathcal{F}$, $O \subset \mathbb{R}$ form a subbase for this topology, so a basic element $B$ for a point $x$ is of the form:

$$B= \cap_{i=1}^{n} (f_i)^{-1}[O_i], \text{ where } n \in\mathbb{N}, f_i \in \mathcal{F}, O_i \subset \mathbb{R} \text{ open.}$$

Now if we had two points $x$ and $y$ that could not be separated by some member from $ \mathcal{F}$, i.e. $\forall f \in \mathcal{F}: f(x)=f(y)$, then for any $B$ of the above form, where $x \in B$, we know that for all $1\le i \le n$: $f_i(y) = f_i(x) \in O_i$ so $y \in B$ and vice versa. So for all open sets $O$ in this initial topology we have $x \in O \leftrightarrow y\in O$. So the space can then not be Hausdorff as witnessed by $x$ and $y$. So by contraposition, if $X$ is Hausdorff some function from $ \mathcal{F}$ assumes different values on $x$ and $y$.

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Henno Brandsma
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