$(X,\tau)$ will be a $T_0$-space but not a $T_1$-space

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?

• 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. – Dog_69 Sep 28 '18 at 22:43

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 $$T_2$$.

$$\tau =\{X,\{\},\{1\}\}$$ or $$\tau =\{X,\{\},\{0\}\}$$

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.