# Definition of a topological space

This may be a simple question, but I am nonetheless confused. I am just starting to learn about topology.

I am confused by part of the Wolfram Mathworld article on Topological Spaces. In it, it is said that a topological space is a collection of open subsets $$T$$ on a set $$X$$ has several essential characteristics.
Two of these make sense to me: the empty set $$\phi$$ is in $$T$$, and that $$X$$ is also in $$T$$. $$\phi$$ because $$\phi$$ is a subset of every set, and $$X$$ because $$T$$ is a collection of subsets of $$X$$.

However, it is then said:

The intersection of a finite number of sets in T is in T.

The union of an arbitrary number of sets in T is in T.

And that if we redefined $$T$$ as a collection of closed subsets, that, with the following changes, we would still have a topological space:

The intersection of an arbitrary number of sets in T is also in T.

The union of a finite number of sets in T is also in T.

I don't understand how the changes from a collection of open to closed subsets, along with the swapping between arbitrary and finite, give us a topological space. Any insight would be appreciated.

It's not too bad. Remember, an open set $U$ yields a closed set $U^c$, by definition, vice versa. Also remember De Morgan's laws: $\left( \cup_{\alpha \in A} U_\alpha \right)^c = \cap_{\alpha \in A} U_\alpha^c$ and $\left( \cap_{\alpha \in A} U_\alpha \right)^c = \cup_{\alpha \in A} U_\alpha^c$

Using this the "open set" set of conditions implies the "closed set" of conditions, and vice versa. You get "finite" and "arbitrary" flipped because De Morgan's laws flip unions with intersections.

More specifically, here's the conversion of the first statement in the first box using quantifiers. Let $O$ be the set of all open sets in $X$.

First step. Require that for any collection of sets $\{U_\alpha\}_{\alpha \in A}$ with $U_\alpha \in O$ then $\cup_{\alpha \in A}U_\alpha \in O$. Second step. Require that for any collection of sets $\{K_\alpha\}_{\alpha \in A}$ with $K^c_\alpha \in O$ then $\cup_{\alpha \in A} K^c_\alpha \in O$. Last step. Use De Morgan's: Require that for any collection of sets $\{K_\alpha\}_{\alpha \in A}$ with $K^c_\alpha \in O$ then $\left(\cap_{\alpha \in A} K_\alpha \right)^c \in O$.

This of course means $\cap_{\alpha \in A} K_\alpha$ is closed knowing that $K_\alpha$ is closed.

• Apologies if this is obvious, but how do I know that $X$ is in $\{K_\alpha\}$? – learner Jan 9 '14 at 19:28
• $X$ may or may not be in the union. What is required is that $X \in O$, that is the set of open sets. – abnry Jan 9 '14 at 19:30
• Yes, sorry, please see my edit for notation. If the collection of sets $\{K_\alpha\}$ is our new $T$, how do we know $X$ is in it - that is, that we still have a topological space? Or am I missing something more fundamental? – learner Jan 9 '14 at 19:32
• Good question. $\emptyset^c = X \in O$, so that means $\emptyset$ is closed. Similarly, $X^c = \emptyset \in O$, so $X$ is also closed. (I am calling $T$ here $O$.) – abnry Jan 9 '14 at 19:34

Note that in topology also as set is called closed iff its complement is open. Therefore, it is merely a switch of perspecitve whether one requires that the intersection of finitely many open sets is open again (where "open" is just a shorthand for "is one of the subsets contained in $T$ of the first kind") or that the union of finitely many of closed sets is closed again (where "closed" is just a shorthand for "is one of the subsets contained in $T$ of the second kind"); this is just deMorgans law if you care to check. Likewise for arbitrary unions of open/arbitrary intersections of closed sets. Note that the complement of $\emptyset$ is $X$ and vice versa, so that deeming $\emptyset$ and $X$ open is equivalent to deeming them closed.

If you present a topological space giving the family of open sets then closed sets are complementary of open sets: i.e. if $X$ is a set and $\tau \subseteq \mathcal P(X)$ is a topology (a family of subsets satisfying the axioms you've written above) then a subset $C \subseteq of X$ is closed if and only if $X \setminus C \in \tau$.

Since the set of closed subset is the set of complementary of open sets it's closed for arbitrary intersection and finite intersection, and of course it contains the empty set and the whole space. That's is because if $\{C_i\}_{i \in I}$ is an indexed family of closed sets then $\bigcap_{i \in I} C_i = X \setminus (\bigcup_{i \in I} (X \setminus C_i))$, so the intersection of closed set is the complementary of the union of arbitrary of open sets(hence complementary of an open set).

Similarly for $C_1,\dots,C_n$ closed sets we have $C_1 \cup \dots \cup C_n= X \setminus ((X \setminus C_1) \cap \dots \cap (X \setminus C_n))$, so a finite union of closed set is the complementary of a finite intersection of open sets, hence is closed.

On the other end for symmetry if you take a set $X$ and family of subsets verifying the property of closed subsets the family of the complementary of these closed sets form a topology (i.e. a family of open set) for $X$.