[Math] Basis for a topology with a countable number of sets

general-topology

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.

Thanks, Adhvaitha

Best Answer

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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