Prove that the topology of a topological space (X, T) is the discrete one if and only if each point of X is open. [duplicate]

Solution 1:

Proof writing is as obvious as you think it is, just need to write it down formally

($\implies$) Assume $\mathcal{T}$ is the discrete topology. Then every subset $S$ of $X$ is open, ie. $S \in \mathcal{T}$. This includes all sets containing just a singleton, ie. all sets of the form $\{ x \}$ for $ x\in X$

($\impliedby$) Assume $\{ x \} \in \mathcal{T}$ for every $ x\in X$. Then notice that for any subset $A \subseteq X$, $A$ has (possibly infinite) elements who are members of $X$, namely $$ A = \{ x_i | x_i \in X \mbox{ and } i\in I \mbox{ where } I \mbox{ is an indexing set} \} $$ But this is exactly a union of singleton sets, $$ A = \bigcup_{i\in I} \{ x_i \} $$ which is open as the union of open sets.

Finally, if $A=\emptyset$ then it open as the intersection of 2 different open sets, ie. $\emptyset = \{ x_1 \} \cap \{ x_2 \}$ where $x_1 \neq x_2$.

If $A=X$, it is simply the union of all singleton sets of elements in $X$ hence open

Solution 2:

If $X$ has discrete topology, then all subsets are open, specially, for any $x\in X$, the singleton $\{x\}$ is open.

If all singletons $\{x\}\subset X$ are open, then any subset $A$ of $X$ is open too since $$A=\bigcup_{x\in A}\{x\}$$ is an arbitrary union of open sets.