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Let $(X, \mathcal T)$ be a $T_1$-space. Now for every $x \in X$, we know that $\{x \}$ is closed in $X$, that is $X \setminus \{x\}$ is open. Consider any subset $S \subseteq X$.

We have that $\displaystyle S = \bigcup_{x \in S} \{x\}$ is closed, since the finite union of closed sets is closed. Now $$X\setminus S = X\setminus \left(\bigcup_{x \in A} \{x\} \right) = \bigcap_{x \in S} \left(X \setminus \{x\}\right),$$ which is the intersection of open sets. That is, $S$ is saturated.

Is this correct? I am having serious doubts about the fact that $S$ is closed, since I'm not too sure $\displaystyle \bigcup_{x \in S} \{x \}$ is in fact closed (since I don't really know that $S$ is finite). Is there any way around this?

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    $\begingroup$ So $S$ is finite? You did not mention this when introducing $S$. $\endgroup$ – Friedrich Philipp Mar 23 '16 at 12:47
  • $\begingroup$ What you actually showed is that the complement of a finite set is saturated. $\endgroup$ – Friedrich Philipp Mar 23 '16 at 12:49
  • $\begingroup$ @FriedrichPhilipp . That is exactly my problem. I don't know if I can assume $S$ to be finite? If I bring in that assumption, will this "proof" still be valid, since what then happens when $S$ is not finite? $\endgroup$ – user290425 Mar 23 '16 at 12:49
  • $\begingroup$ Look at your main line. It shows that each complement is the intersection of open sets, right? $\endgroup$ – Friedrich Philipp Mar 23 '16 at 12:51
  • $\begingroup$ @FriedrichPhilipp that is correct. $\endgroup$ – user290425 Mar 23 '16 at 12:52
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Let $S\subset X$ and let $S^c := X\setminus S$. Then $$ S = X\setminus S^c = X\setminus\bigcup_{x\in S^c}\{x\} = \bigcap_{x\in S^c}(X\setminus\{x\}). $$ Since the singleton sets are closed (due to $T_1$), this is an intersection of open sets.

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  • $\begingroup$ Just for interest sake - if for some topological space $(X,T)$ we have that every subset $S$ is saturated, then can we deduce that $(X,T)$ is a $T_1$-space? I am lead to believe yes, since: Suppose $S\subseteq X$ is saturated for any subset $S$. Now , the same as in the above proof we have that \begin{align}S = \bigcap_{x \in S^C} \left( X \setminus \{ x \} \right)\end{align} But since we know $S$ is saturated, we must have that $X\setminus \{x \}$ is open, that is $\{x \}$ is closed. Hence $(X,T)$ is a $T_1$-space. $\endgroup$ – user290425 Mar 23 '16 at 13:40
  • $\begingroup$ No. Your conclusion that $X\setminus\{x\}$ is open is false. The open sets whose intersection is $S$ might be totally different from the $X\setminus\{x\}$. $\endgroup$ – Friedrich Philipp Mar 23 '16 at 13:55
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    $\begingroup$ However, the claim is correct. Let $x,y\in X$, $x\neq y$. Since $\{x\}$ is saturated, we find open sets $U_i$, $i\in I$ (where $I$ is some index set) such that $\{x\} = \bigcap_i U_i$. Since $y\neq x$, there exists some $U_x = U_{i_0}$ such that $y\notin U_x$. So, we have found a neighborhood of $x$ which does not contain $y$. Similarly, we find a nbh of $y$, which doesn't contain $x$. So, $X$ is $T_1$. $\endgroup$ – Friedrich Philipp Mar 23 '16 at 14:00

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