# Basis for a topology with a countable number of sets

I am working on the problems from the textbook "Topology without tears". I am stuck with problem number $4$ in Exercise $2.2$. Could anyone suggest some hints on how to proceed? The question goes as follows.

A topological space $(X,\tau)$ is said to satisfy the second axiom of countability or to be second countable if there exists a basis $\mathcal{B}$ for $\tau$, where $\mathcal{B}$ consists of only a countable number of sets.

1. Prove that the discrete topology on an uncountable set does not satisfy the second axiom of countability.

2. Let $(X,\tau)$ be the set of all integers with the finite-closed topology. Does the space $(X,\tau)$ satisfy the second axiom of countability?

For the first problem, I tried to argue by contradiction assuming that the basis for the topology is countable. But I did-not know what I need to look at to prove the contradiction. For the second one, I do not know where to start. I have not thought deeply on the second one though.

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For the first problem, show that for each $x\in X$, we must have $\left\{x\right\}\in\mathcal B$. For the second question, remember that the collection of the finite subsets of $\mathbb N$ (or any countable set) is countable. –  Davide Giraudo Sep 4 '11 at 18:11
Yes. Thats what I wanted to show for the first one but I do not know how to proceed i.e for the first one we must have (\{x\} \in \mathcal{B}) for every (x \in X). But how do I proceed to show that if the basis is a countable set? –  Adhvaitha Sep 4 '11 at 18:15
$\left\{x\right\}$ is open since the topology is discrete, and $\left\{x\right\}$ should be an union of sets of the basis. Each set of this union consists at most of one element (namely $x$) an one of these sets should be $\left\{x\right\}$. –  Davide Giraudo Sep 4 '11 at 18:36
@Adhwaitha: Every $1$ element set is open. Every open set is a union of elements of $\mathbb{B}$. Therefore every $1$-element set is in $\mathbb{B}$. But there are uncountably many $1$-element sets, since we are dealing with an uncountable set. Therefore $\mathbb{B}$ cannot be countable (a countable set cannot have an uncountable subset). –  André Nicolas Sep 4 '11 at 18:42

The first question can be proved directly:

Since for every $x$ we have that $\{x\}$ is open, and in every basis for the topology every open set should be a union of basic open sets, the singletons cannot be anything but the union of themselves. Therefore every basis for the topology must contain $\Big\{\{x\}\mid x\in X\Big\}$ which is uncountable since it has the same cardinality of $X$ itself.

The second question can be easily solved with a theorem from elementary set theory:

Theorem: Suppose $X$ is a countable set, then $\{B\subseteq X\mid B\text{ finite}\}$ is also countable.

Proof: Fix some enumeration of $X=\{x_n\mid n\in\mathbb N\}$. Suppose $A\subseteq X$ is a finite set, define: $$f(A) = \sum_{x_i\in A} 2^i$$

(In the $\mathbb N$ case, encode every finite set as a binary number)

The proof that this function is a bijection is done by strong (complete) induction, and is not difficult.

Now we can proceed with the second question. From the theorem, since the integers are countable, there are only countably many finite subsets of integers.

Since every set is closed if and only if it is finite, there are countably many finite closed sets, and therefore finitely many open sets (the complements of closed sets).

So the entire topology is a countable basis for itself, as needed.

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Thanks. That clarified everything. –  Adhvaitha Sep 5 '11 at 13:15
For the point one, my hint is to think about the base $$\mathcal{B}_0:=\{\{x\}|x\in X\}$$ and thinking in its relation to all the other posible basis of the discrete topology on a set $X$. If $\mathcal{B}$ is a basis for the discrete topology, what is its relation with $\mathcal{B}_0$? (See that for each $x\in X$, $\{x\}$ is open, and there is a nonempty open set $U\in\mathcal{B}$ such that $x\in U\subseteq \{x\}$.) And what is the cardinality of $\mathcal{B}_0$? (Clearly, $\mathcal{B}_0$ is equipotent to $X$.)
For proving that, my clue is to center on finite sets of order $n$. And how there is an injective map from the set of finite sets of order $n$ to the cartesian product of $X$ with itself $n$ times.
Thanks. As I have mentioned in the comments below the question, I wanted to show for the first one we must have $\{x\} \in \tau(\mathcal{B})$ for every $x \in X$. But how do I proceed to show that if the basis is a countable set? The basis $\mathcal{B}_0$ is uncountable. But how does it guarantee any other basis is also uncountable? I can see it is true but I am missing the connecting link. For the second one, every subset of $X$ is countable. But how do I proceed to show that the basis for the topology is also countable? –  Adhvaitha Sep 4 '11 at 18:21
@Adhvaitha: The point is that for each $x\in X$, every basis must contain $\{x\}$: $\{x\}$ is open, so it must be a union of basis sets, but it’s a singleton, so this is possible only if it is a basis set. In other words, if $\mathcal{B}$ is any basis for the discrete topology, $\mathcal{B}_0 \subseteq\mathcal{B}$. –  Brian M. Scott Sep 4 '11 at 19:30