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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!

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  • $\begingroup$ $\tau_1,\tau_2$ 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? $\endgroup$ – mb- Sep 10 '16 at 17:36
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    $\begingroup$ I corrected your terminology: you’re intersecting topologies, not spaces. Yes, your example is fine. $\endgroup$ – Brian M. Scott Sep 10 '16 at 17:38
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    $\begingroup$ 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. $\endgroup$ – Marios Gretsas Sep 10 '16 at 17:38
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    $\begingroup$ Your example is perfectly valid. Good job. $\endgroup$ – Crostul Sep 10 '16 at 17:39
  • $\begingroup$ Oh i see.Thank you !:) $\endgroup$ – Marios Gretsas Sep 10 '16 at 17:41
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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. $\square$

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