Smallest topology on $\mathbb{C}$ such that all singletons are closed

general-topologyseparable-spaces

This question was part of an exam for which I am preparing.

Question was related to finding smallest topology T on $\mathbb{C}$ in which all singletons are closed.
and then findind which options are true amog these:

A ($\mathbb{C}$,T) is haursdorff.

B ($\mathbb{C}$,T) is compact.

C ($\mathbb{C}$,T) is connected.

D $\mathbb{Z}$ is dense in ($\mathbb{C}$,T) .

But I am just a beginner in topology and self studying from Wayne patty . I was unable to find what topology should be . So, can you please tell what topology should be . rest I would prefer to work by mysely.

Thank you.

Best Answer

I will answer the questions A and D, as you asked in the comments.

A is false because if we have two points and any the neighborhoods of them, say $U$ and $V$, they must satisfy $|\mathbb C\setminus U|,|\mathbb C\setminus V|<\infty$ and their intersection, $U \cap V$, is an open subset. If $U\cap V$ is empty, then its complement is union of finite sets, that is $\mathbb C= \mathbb C\setminus (U\cap V)=(\mathbb C \setminus U)\cap (\mathbb C\setminus V)$ is finite, an absurd!

For proving D, we take any non empty, open subset $U \subseteq \mathbb C$, we have $|\mathbb C\setminus U|<\infty$. If $\mathbb Z \cap U =\emptyset$, then $\mathbb Z \subseteq \mathbb C\setminus U$. That is, $\mathbb Z$ is finite, another absurd! So, $\mathbb Z \cap U \neq \emptyset$. Therefore, the integers are dense in $\mathbb C$ with that topology (in fact, any infinite set).

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