A question about topology regarding its conditions A question from a rookie.
As we know, $(X, T)$ is a topological space, on the following conditions,


*

*The union of a family of $T$-sets, belongs to $T$;

*The intersection of a FINITE family of $T$-sets, belongs to $T$;

*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.

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.
