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Consider the following topology on three point set $X$=$\{a,b,c\}$. $\tau=\{X,\emptyset,\{a\},\{a,b\}\}$.

This is clearly a topology on $X$.

Can somebody please explain me how this becomes a $T_0$ space and not a $T_1$ space?

For the two points $a$ and $c$, we have the open set $a$ in $\{a\}$ and $\{c\}$ is not in $\{a\}$. So is $T_0$. but for the two points $a$ and $b$, $a$ in $\{a,b\}$ and $b$ is also in $\{a,b\}$, so isn't $T_0$. But it says that this is $T_0$. How can this be?

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    Be careful that you're distinguishing between your points and your open subsets.2012-11-11

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A space $X$ is $T_0$ if whenever $x$ and $y$ are distinct points, at least one of them has an open neighborhood not containing the other. It’s not necessary that each of them have an open neighborhood not containing the other. In fact, if that happens, the space is $T_1$.

In this space $\{a\}$ is an open set containing $a$ but not $b$ or $c$, which takes care of the pairs $\{a,b\}$ and $\{a,c\}$; $\{a,b\}$ is an open set containing $b$ but not $c$, which takes care of the one remaining pair of points, $\{b,c\}$.

However, $X$ is not a $T_1$ space because (for instance) $b$ has no open neighborhood that does not contain $a$, as you saw. For that matter, $c$ has no open nbhd that does not contain $a$ and $b$, so the space fails very badly to be $T_1$.

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    @ccc: Great! You’re welcome.2012-11-11