# Is the intersection of $T_0$ topologies a $T_0$ topology?

A topological space $(X,\tau)$ is a $T_0$ space (or $Kolmogorov$ space), if for every two distinct elements $x,y$$\in X we can find an open set that contains x and not y or an open set that contains y and not x. Is the intersection of T_0 topologies a T_0 topology? My proof to this: Let X=\{a,b\} and \tau_1,\tau_2 two topologies on X with \tau_1=\{X,\{a\},\emptyset\} and \tau_1=\{X,\{b\},\emptyset\}. This two topologies on X form two T_0 spaces. Then \tau_1\cap\tau_2$$=\{X,\emptyset\}$ which is the trivial topology on $X$.

But we can easily prove that a space with the trivial topology is not a $T_0$ space.

Is this an adequate counterexample for the above statement?

Thank you in advance!

• $\tau_1,\tau_2$ is not $T_0$ spaces, as you cannot find an open set which contains $b$ but doesn't contain $a$ in the first case, and vice versa in the second? – mb- Sep 10 '16 at 17:36
• I corrected your terminology: you’re intersecting topologies, not spaces. Yes, your example is fine. – Brian M. Scott Sep 10 '16 at 17:38
• in the definision of a To space it suffices to find an open set that does not contain the other of the two element OR the other way around. I believe so. – Marios Gretsas Sep 10 '16 at 17:38
• Your example is perfectly valid. Good job. – Crostul Sep 10 '16 at 17:39
• Oh i see.Thank you !:) – Marios Gretsas Sep 10 '16 at 17:41

## 1 Answer

Yes, that counterexample works! To spell out in detail:

For $\tau_1$, the open set $\{a\}$ establishes that $\tau_1$ is Kolmogorov, since the only two points are $a$ and $b$.

Analogously for $\tau_2$, the open set $\{b\}$ establishes that $\tau_2$ is Kolmogorov.

The intersection $\tau_1 \cap \tau_2 = \{\varnothing, \{a,b\}\}$ is not Kolmogorov, because every open set which contains $a$ also contains $b$, and vice-versa. $\square$