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Exercise: A topological space is said to be $T_0$-space if for each pair of distinct points $a,b$ in $X$, either there exists an open set containing $a$ and not $b$, or there exists an open set containing $b$ and not $a$.A topological space $(X,\tau)$ is said to be $T_1$-space if every singleton set $\{x\}$ is closed in $(X,\tau)$

Put a topology on the set $X=\{0,1\}$ so that $(X,\tau)$ will be a $T_0$-space but not a $T_1$-space.

If I consider the topology $\tau=\{X,\emptyset,\{0\}\}$, the point $0\in\{0\}$ and $\{0\}\in X$. If I come up with other topology since I need to have $X$ in the topology there will be always two open sets containg the same point.

Question:

So how do I build a topology on this case that $(X,\tau)$ will be a $T_0$-space but not a $T_1$-space? What am I thinking wrong?

Thanks in advance!

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  • $\begingroup$ Just a remark. No matter if there are two (or more) open sets containing the same point. The important is that there's at least one that contains $a$ and does not contain $b$. Think in $R$. There are many open sets that contain the $0$, for instance open intervals $(-n,n)$. But given any other point $x\in R$ you can find an open set that contains $0$ and not $x$: the open interval $(-|x|/2,|x|/2)$, for example. $\endgroup$ – Dog_69 Sep 28 '18 at 22:43
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I'm not sure what your doubt is. The family $\tau=\{\emptyset,\{0\},X\}$ is a topology on $X=\{0,1\}$.

It is $T_0$, because there is an open set containing $0$ and not $1$, precisely $\{0\}$. Since these are all the points in $X$, we are done.

On the other hand, $\{0\}$ is not closed, because the closed sets are $X$, $\{1\}$ and $\emptyset$. Thus the topology is not $T_1$.


There are not many choices for a topology on $X=\{0,1\}$: the following is the complete list.

  1. $\{\emptyset,X\}$, the trivial topology; not $T_0$.
  2. $\{\emptyset,\{0\},X\}$; it is $T_0$ but not $T_1$.
  3. $\{\emptyset,\{1\},X\}$; it is $T_0$ but not $T_1$.
  4. $\{\emptyset,\{0\},\{1\},X\}$, the discrete topology; it is $T_2$.
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$\tau =\{X,\{\},\{1\}\}$ or $\tau =\{X,\{\},\{0\}\}$

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Not every singleton can be closed or $X$ would be $T_1$. So pick e.g. $\{1\}$ to not be closed. Then $\{0\}$ cannot be open, as its complement. This means the only open set that contains $0$ must be $\{0,1\} = X$. But $\{1\}$ must be open for $X$ to be $T_0$ (we must distinguish $1$ from $0$). $X$ and $\emptyset$ are always open.

So we get $\{\emptyset, X , \{1\}\}$ as the topology (we can also choose $0$ to be the open singleton, and we get a different, but essentially the same, topology). It's called the Sierpiński two-space.

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