Showing that a topological space is ${\rm T}_1$

Question

Let $X$ be a topological space and let $\Delta = \{(x,x) : x\in X \}$ be the diagonal of $X\times X$ (with the product topology).

I was asked to prove that $X$ is ${\rm T}_1$ if and only if $\Delta$ can be written as intersection of open subsets of $X\times X$.

I think is better (maybe easier) to use the well-known result "$X$ is ${\rm T}_1$ if and only if $\{x\}$ is a closed set $\forall x\in X$".

What I've done:

Assuming that $X$ is ${\rm T}_1$, let $Y=(X\times X) \setminus\Delta$ and note that (trivially) $$Y = \bigcup_{y\in Y} \{y\}.$$ Then $$\Delta = (X\times X) \setminus Y = (X\times X) \setminus \left(\bigcup_{y\in Y} \{y\} \right) = \bigcap_{y\in Y} (X\times X)\setminus \{y\},$$ where $(X\times X)\setminus\{y\}$ is open since each $\{y\}$ is closed.

The other direction seems be a bit harder, I've tried unsuccessfully. Any ideas?

Thanks in advance.

Answer 1

Suppose that

$$\Delta = \bigcap \{O_i: i\in I\}$$ for some index set $I$, and all $O_i$ open in $X \times X$.

Let $x\neq y$, to show $T_1$-ness, we need to find an open set $O$ of $X$ that contains $x$ but misses $y$. So consider $(x,y)$, which is not in $\Delta$, so there is an $i \in I$ such that $(x,y)\notin O_i$.

Now $(x,x) \in \Delta \subseteq O_i$, so by using the product base we can find an open set of the form $U\times V$ that contains $(x,x)$ and sits inside $O_i$.

But then $O=U \cap V$ contains $x$ but not $y$, as $y \in U \cap V$ would imply $(x,y) \in U \times V \subseteq O_i$, while we had $(x,y) \notin O_i$ by choice of $i$. So we have our open set as required.

Answer 2

Just an idea.

Suppose $\Delta=\bigcap O_\lambda$, where $(O_\lambda)_{\lambda\in\Lambda}$ is a family of open sets in $X\times X$. Let $(x,y)\in X\times X$ with $x\neq y$. Then $(x,y)\notin \Delta$ and therefore $(x,y)\notin\bigcap O_{\lambda}$. So choose $\mu\in\Lambda$ such that $(x,y)\notin\ O_\mu$; say $O_\mu=\pi_1^{-1}(U_\mu)\cap\pi_2^{-1}(V_\mu)$ where $U_\mu,V_\mu$ are open in $X$. Then $(x,y)\notin \pi_1^{-1}(U_\mu)$ or $(x,y)\notin \pi_2^{-1}(V_\mu)$. Assume $(x,y)\notin \pi_1^{-1}(U_\mu)=U_\mu\times X$. Then $x\notin U_\mu$ as $U_\mu$ is open. But $(x,x)\in\Delta\subseteq O_\mu\subseteq \pi_1^{-1}(U_\mu)=U_\mu\times X$; so $x\in U_\mu$. And if you assume the latter case then you'll get $y\notin V_\mu$ and $y\in V_\mu$.

Is something wrong?

Answer 3

If $X$ is not $T_1$ let $x,y\in X$ with $x\ne y\in \overline {\{x\}}$.

Let $F$ be any family of open subsets of $X^2$ such that $\cap F\supset \Delta.$ For each $f\in F$ there exists open $U_f$ and $V_f$ of $X$ such that $(y,y)\in U_f\times V_f\subset f.$

But $x\in U_f$ because $y\in U_f$ and $y\in \overline {\{x\}}$. Also $y\in V_f.\;$ So $(x,y)\in U_f\times V_f\subset f .$ As this holds for every $f\in F ,$ we have $(x,y)\in\cap F.$

But $(x,y)\not \in \Delta,$ so $\cap F \ne \Delta.$

Answer 4

Instead of tackling the OP's immediate question, we start with no assumptions and consider any topological space $X$.

Lemma: Let $u$ and $v$ be two distinct points belonging to $X$, The following two statements are equivalent:

(1) Every open set of $u$ contains $v$.

(2) Every open set in the product space $X \times X$ containing $(u,u)$ must also contain $(v,u)$.

Proof:

As you construct open set in the generating base of $X \times X$ containing $(u,u)$, you will see that $(v,u)$ always 'comes for the ride' in the open 'box'. It works both ways:

(1) iff (2) iff a point 'comes for the ride'

QED

Note that the definition of a T1 space can be viewed thru the lens of 'negating' the Lemma's premises and conclusions. The NOT-(1) side is the definition and the NOT-(2) is an equivalent formulation.

We now address the OP's question concerning the 'other direction'.

Assume that Δ can be written as intersection of open subsets. We will show that the negation of Lemma-(2) holds to show that $X$ is T1.

Let $u$ and $v$ be any two distinct points so that $(v,u)$ is not on Δ. Let $U$ be an open set in $X \times X$ containing Δ such that $(v,u) \notin U$. The existence of this $U$ shows that $X$ is a T1-space.