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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. –  Cameron Buie Nov 11 '12 at 6:34

1 Answer 1

up vote 2 down vote accepted

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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@ Brian,what did you mean by"....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}"? –  ccc Nov 11 '12 at 6:20
2  
@ccc: The definition of $T_0$ says that for each pair of distinct points in the space, at least one of them must have an open nbhd that misses the other. In this space there are only three pairs of distinct points, $\{a,b\},\{a,c\}$, and $\{b,c\}$. For each of those pairs I found an open set containing one point but not the other; the open set $\{a\}$ works for the first two pairs, and the open set $\{a,b\}$ works for the third pair. That’s all that’s required to show that the space is $T_0$. –  Brian M. Scott Nov 11 '12 at 6:30
    
oh I see!!!now only I understood that..:)thanks alot Brian!!:) –  ccc Nov 11 '12 at 6:33
    
@ccc: Great! You’re welcome. –  Brian M. Scott Nov 11 '12 at 6:35

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