Let $X$ be a set such that $\mathbf{card}(X) \geq 2$ and $\tau_x = \{ \emptyset, X \}$. Let $(X,\tau_x)$ be a topological space. I have to show that $(X,\tau_x)$ is regular and normal.

A topological space is regular if $\forall x \in X$, $F \in \tau^{c}_x$, $x \notin F, \exists U_x, U_F \in \tau_x$ such that $x \in U_x, F \subseteq U_F, U_x \cap U_F = \emptyset.$

This is what I've done: $\forall x \in X, \emptyset \in \tau^{c}_x$ and $x \notin \emptyset$. Then, $\exists U_1, U_2 (U_1 = X, U_2 = \emptyset) \in \tau_x$ such that $x \in U_1 = X, \emptyset \subseteq U_2 = \emptyset$ and $U_1 \cap U_2 = \emptyset$, which is true because $X \cap \emptyset = \emptyset$.

A topological space is normal if $\forall F_1, F_2 \in \tau^{c}_x, F_1 \cap F_2 = \emptyset, \exists U_1, U_2 \in \tau_x$ such that $F_1 \subseteq U_1, F_2 \subseteq U_2, U_1 \cap U_2 = \emptyset$.

This is what I've done: As $\tau_x = \{ \emptyset, X \}$, then $\tau^{c}_x = \{ \emptyset, X \}$. The only pair of closed sets of $\tau^{c}_x$ are $\emptyset$ and $X$. Indeed, $X \cap \emptyset = \emptyset$. Then, $\exists U_1, U_2 (U_1 = X, U_2 = \emptyset) \in \tau_x$ such that $F \subseteq U_1 = X, \emptyset \subseteq U_2 = \emptyset$ and $U_1 \cap U_2 = \emptyset$, which is true because $X \cap \emptyset = \emptyset$.

Am I missing something on my proofs? Also, I don't understand why $\mathbf{card}(X)$ must be $\mathbf{card}(X) \geq 2$.

  • $\begingroup$ It does not have to be. Both statements are true for any $X$. $\endgroup$ – freakish Apr 18 '18 at 15:24

The proofs are essentially correct: we just note that for the trivial topology, the only way we can have $x \notin F$ and $F$ closed, is when $F=\emptyset$, and then we take $\emptyset$ as the neighbourhood of $F$ and $X$ for $x$.

Also, the only pair of disjoint closed sets are $\emptyset$ and $X$ and they have themselves as open disjoint neighbourhoods as well.

So they are normal / regular for trivial reasons. No need to write down formulae and formal arguments as you did. What I wrote above summarises that in more plain language, which is often better.

$|X| \ge 2$ is not really needed as such. The one-point space and the empty space are also normal/regular for similarly trivial reasons.


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