A question about topology regarding its conditions

Question

A question from a rookie.

As we know, $(X, T)$ is a topological space, on the following conditions,

1. The union of a family of $T$-sets, belongs to $T$;
2. The intersection of a FINITE family of $T$-sets, belongs to $T$;
3. The empty set and the whole $X$ belongs to $T$.

So when the condition 2 is changed into:

2'. The intersection of a family of $T$-sets, belongs to $T$;

can anyone give a legitimate topological space as a counterexample to condition 2'?

Thank you.

Answer 1

The intersection of intervals $(-1/n,\ 1/n)$ for $n\in\mathbb N$ is only $\{0\}$ which is not open in $\mathbb R$ with the euclidean topology.

Answer 2

Stefan H. has provided perhaps the canonical counterexample. But just for some further information, a topological space in which the intersection of arbitrary families of open sets are open is called an Alexandrov space. A basic fact is that a topological space $X$ is an Alexandrov space iff every $x \in X$ has a smallest open neighbourhood.

In the ($\Rightarrow$) direction, if $X$ is an Alexandroff space, then for each $x \in X$ the intersection of all open neighbourhoods of $x$ will itself be an open neighbourhood of $x$, and will be a subset of all open neighbourhoods of that point. In the ($\Leftarrow$) direction, if by $U_x$ we denote the smallest open neighbourhood of $x \in X$, then given any family $\mathcal{U}$ of open sets, if $x \in \bigcap \mathcal{U}$, then $U_x \subseteq U$ for all $U \in \mathcal{U}$, and so in fact $U_x \subseteq \bigcap \mathcal{U}$: it follows that $\bigcap \mathcal{U} = \bigcup \{ U_x : x \in \bigcap \mathcal{U} \}$, and is an open set.

Thus if any point in a topological space does not have a smallest open neighbourhood, the space is not Alexandrov.

In the presence of even a weak separation axiom like T$_1$-ness, Alexandrov spaces are quite simple: they are all discrete! Therefore any non-discrete T$_1$-space is not Alexandrov.