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Could you help me to prove this proposition below?

$X$ is a finite set and $(X,\tau)$ is a topological space. Show that if every subset which have single element is closed set then $\tau$ describes discrete topology.

I started with let $X=\{x_1,x_2,\cdots,x_n\}$ . If every subset with single element is closed set then for $1\leq i \leq n$ then $\{x_i\}$ is closed, at the same time $X \setminus \{x_i\}$ is open set.

for$\quad$ $1\leq i \leq n$ $\quad$ $X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ and we know that $(X,\tau)$ is a topological space so it provides the axioms. It is the point where i stuck. How can i build discrete topology with using the set $\quad$ $X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ by using finite intersection and (finite or infinite) union set operations. thank you in advance :)

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3 Answers 3

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Since the union of finitely many closed sets is closed, and in a finite set every subset is a finite union of singletons, it follows that every subset is closed. Therefore it follows that the complement of every set is closed, and so every set is open. Therefore we have the discrete topology.

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Consider an intersection of two such sets, \begin{align*} (X \setminus \{ x_i \}) \cap (X \setminus \{ x_j\}) &= X \cap \{ x_i \}^c \cap X \cap \{ x_j \}^c = X \cap \{ x_i, x_j \}^c \\ &= X \setminus \{ x_i, x_j\}\,. \end{align*} Now iterate the procedure to generate all the open sets you need.

In fact it is easier to work with properties of closed sets. If $\{ x_i\}$ and $\{ x_j\}$ are both closed, then so is $\{ x_i\} \cup \{x_j\} = \{x_i, x_j\}$.

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Call $\{1,..,n\}=I$. As you wrote, $A_i = X \setminus \{x_i\}=\{x_1,x_2,\cdots,x_{i-1},x_{i+1},\cdots,x_n\}$ is open. If you take the finite intersection of $\bigcap_{i \in I \setminus \{j\} }A_i$, you get an $\{ x_j \}$.

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