Closure as a map

Question

Apparently the following two definitions of a topology are equivalent:

(A) Via a map $\tau: \mathcal{P}(X) \rightarrow \mathcal{P}(X)$ such that: $$ \quad \forall S \in \mathcal{P}(X), S \subseteq \tau(S) \qquad [1] \\ \tau\bigg( \underset{S \in \Sigma}{\bigcup} S\bigg) = \underset{S\in\Sigma}{\bigcup} \tau(S) \qquad [2] \\ \tau(\tau(S)) = \tau(S) \qquad [3] $$

(B) Via a definition of 'closeness',where:
$$ x \; \text{is close to} \; S \in \mathcal{P}(X) \iff x \in \tau(S) $$ and:

• i) If $x \in S$, $x$ is close to $S$

• ii) For any finite collection of subsets $\{S_{1}...S_{n}\}$ if $x$ is close to $S_{1} \cup... \cup S_{n}$, $\exists 1\le i \le n$ such that $x$ is close to $S_{i}$

• iii) If $S$ and $T$ are two sets such that every $s \in S$ is close to $T$, then if $x$ is close to $S$ it is also close to $T$

Attempt to show equivalence:

• $i)$ and $[1]$ are clearly the same, as are $ii)$ and $[2]$. But why does $[3] \iff iii)$? $iii)$ is just the statement:

$$ S \subseteq \tau(T) \implies \tau(S) \subseteq \tau(T) $$, but I can't see a way to make this that equivalent to $\tau(\tau(S)) = \tau(S)$? If $ V = \tau(S)$, then $S \subseteq \tau(V)$ $( S \subseteq \tau(S) \subseteq \tau(V)$). So $\tau(S) \subseteq \tau(\tau(S))$ [or by $2$] - but how do we prove $\supseteq$, and why does $\tau(\tau(S)) = \tau(S) \implies iii)$?

Thanks!

Answer 1

The second axiom(s) (which only applies to finite unions!) implies the lemma that $A \subseteq B \rightarrow \tau(A) \subseteq \tau(B)$, as $A \subseteq B$ implies $B = A \cup B$ and so $\tau(B) = \tau(A) \cup \tau(B)$ which implies $\tau(A) \subseteq \tau(B)$, as claimed.

So from (i) we have $S \subseteq \tau(S)$ and thus from the lemma $\tau(S) \subseteq \tau(\tau(S))$. If now (iii) also holds, from $\tau(S) \subseteq \tau(S)$ we can conclude $\tau(\tau(S)) \subseteq \tau(S)$, and we have equality and [3].

If [3] holds, and $S \subseteq \tau(T)$ then the lemma says $\tau(S) \subseteq \tau(\tau(T)) = \tau(T)$ and then (iii) is proved.

Basically you then have the equivalence of two axiom systems for $\tau$, [1],[2],[3] and [1],[2], (iii).

You are still missing the axiom $\tau(\emptyset) = \emptyset$ (or we could define $\tau(A) = X$ for all subsets of $X$ and the result would not be a topology as the empty set is not closed. It's a minor thing, but still.