A sufficient condition for a space not to be $T_1$ by a collection generating the topology.

Question

I've thought about the following result, which I wanted to verify:

Let $X$ be a topological space where $\vert X\vert>1$ generated by a collection of subsets $\{ S_\alpha \}_{\alpha\in \Lambda}$, such that $\cup_{\alpha\in \Lambda} S_\alpha \neq X$, then $X$ is not a $T_1$ space.

My reasoning for this is that if $X$ is a topological space with more than one point, and it has a point $x\in X$ without a proper neighbourhood, then $\overline{ \{x \} }=X$. This follows from the fact that $y\in X$ is in $\overline{E}$ if and only if $U\cap E\neq \emptyset$ for any neighbourhood $U$ of $y$.

Is this logic sound? I am currently trying to deduce conditions from a generating collection about whether it does not satisfy separation axioms, so if there is a similar stronger result I would be happy to be informed of it.

Answer 1

My reasoning for this is that if $X$ is a topological space with more than one point, and it has a point $x\in X$ without a proper neighbourhood, then $\overline{ \{x \} }=X$.

Let's see, consider $X=\{1,2\}$ with topology $\tau=\big\{\emptyset,\{1\},\{1,2\}\big\}$. Note that $(X,\tau)$ satisfies your assumption, since it is generated by $\big\{\{1\}\big\}$. Clearly the only open neighbourhood of $2$ is $X$ itself. But $\{2\}=X\backslash\{1\}$ is closed, i.e. $\overline{\{2\}}=\{2\}$.

It's the other way around: under your assumption the only point that has a chance of being closed is the special one you've chosen.

Assume that $x_0\in X$ is a point such that $X$ is the only open neighbourhood of $x_0$ (such point exists by your condition). It follows that $x_0$ belongs to the closure of any subset of $X$. In particular $x_0\in\overline{\{x\}}$ for any $x\in X$. Thus almost all (possibly except $x_0$) points are not closed. Hence $X$ is not $T_1$.

Answer 2

Assuming the given hypothesis, it's true that $X$ is not $T_1$, but your argument is not correct.

You claimed: If some $x\in X$ does not have a proper neighborhood, then $\overline{\{x \} }=X$.

As a counterexample, let $X$ be any set with $|X| > 1$, and for some fixed $x\in X$, let the topology on $X$ be such that the proper open subsets of $X$ are those sets not containing $x$. Then $x$ has no proper neighbourhood, but the complement of $\{x\}$ is open, hence $\overline{\{x\}}=\{x\}\ne X$.

To prove the statement in question, let $V$ be the union of the elements of $S$.

Then every proper open subset of $X$ is a subset of $V$.

By hypothesis, $V\ne X$.

Consider two cases . . .

Case $(1)$:$\;V\ne{\large{\varnothing}}$.

Let $a\in V$ and let $b\in X{\setminus}V$.

Then $b\in X{\setminus}\{a\}$ and $b\not\in V$, hence $X{\setminus}\{a\}$ is a proper open subset of $X$, but not a subset of $V$.

Hence $X{\setminus}\{a\}$ is not open, so $\{a\}$ is not closed.

Case $(2)$:$\;V={\large{\varnothing}}$.

Let $a\in X$.

By hypothesis, we have $|X| > 1$, so $X{\setminus}\{a\}\ne{\large{\varnothing}}$.

Then $X{\setminus}\{a\}$ is a nonempty proper subset of $X$, hence $X{\setminus}\{a\}$ is not open, so $\{a\}$ is not closed.

In either case, $X$ has a singleton subset which is not closed, hence $X$ is not $T_1$.