Sequence in first-countable compact topological space

[Monocles]

Homework Statement

In a first countable compact topological space, every sequence has a convergent subsequence.

Homework Equations

N/A

The Attempt at a Solution

I'm self-studying topology, so I'm mostly trying to make sure that my argument is rigorous. I understanding intuitively why this argument doesn't work for a non-compact topological space, but I don't know why in a rigorous sense. I also have yet to take analysis, so I'm not very confident with sequences. Finally, I don't make use of the fact that the space is first-countable, which makes me suspicious.

Let $$X$$ be a first countable compact topological space, and let $$\{ p_n \}$$ be a sequence in $$X$$. Since $$X$$ is compact, we will pick two open sets $$U_{1_a}$$ and $$U_{1_b}$$ such that $$U_{1_a} \cup U_{2_a} = X$$ and $$U_{1_a}\cap U_{1_b} \neq X$$. By the pigeonhole principle, an infinite number of points in the sequence $$\{ p_n \}$$ will lie within one of these sets. Without loss of generality, we can say that this set is $$U_{1_a}$$. Within $$U_{1_a}$$, we pick one point of $$\{ p_n \}$$ that lies within $$U_{1_a}$$ and call it $$q_1$$ to being constructing a subsequence.

Next, we cover $$U_{1_a}$$ with two open sets $$U_{2_a}$$ and $$U{2_b}$$ such that $$(U_{2_a} \cup U_{2_b}) \cap U_{1_a} = U_{1_a}$$ and $$U_{2_a} \cap U_{2_b} \neq U_{1_a}$$. By the pigeonhole principle, an infinite number of points of the sequence $$\{ p_n \}$$ must lie within at least one of $$U_{2_a}$$ and $$U_{2_b}$$. We will label this set with $$U_{2_a}$$ without loss of generality. We pick a point of $$\{p_n\}$$ that lies within $$U_{2_a}$$ and label it $$q_2$$, declaring it to be the second point of the subsequence.

We consider with this process inductively, to obtain a subsequence $$\{q_n\}$$ that converges to some $$q$$ that lies within $$U_{1_a} \cap U_{2_a} \cap \ldots$$. Q.E.D.

I do not believe that this argument is correct because I do not make use of the fact that the space is first-countable. But, I do not know how to make use of that fact without being given a point that the subsequence converges to! So, where am I going wrong? I also do not think that I am making proper use of the fact that the space is compact, but I could be wrong.

 

[mighty2000]

Thank you for your post. Your argument is on the right track, but there are a few key points missing that make it not quite rigorous. Let me try to guide you through the correct proof.

First, let's clarify the statement we are trying to prove. The statement is that in a first countable compact topological space, every sequence has a convergent subsequence. This means that given any sequence, we can always find a subsequence that converges to some point in the space.

Now, let's consider your approach. You have correctly identified that in a compact space, we can cover the space with finitely many open sets, and at least one of these sets must contain infinitely many points of the sequence. However, your choice of open sets U_{1_a} and U_{1_b} is not quite correct. Remember, we want to find a subsequence that converges to some point in the space. This means that our open sets should be chosen in such a way that they "converge" to a point in the space. In other words, we want to choose open sets that are getting smaller and smaller, and eventually "pinch" the point we want our subsequence to converge to.

This is where the first countability of the space comes into play. Recall that a space is first countable if for every point, there exists a countable basis of open sets containing that point. In other words, we can "zoom in" on any point in the space by choosing smaller and smaller open sets. So, starting with our sequence \{p_n\}, we can choose a point p_1 and an open set U_1 containing p_1. Then, we can choose a point p_2 and an open set U_2 containing p_2, such that U_2 is contained in U_1 (since U_1 is getting smaller and smaller). Continuing in this way, we can construct a subsequence \{q_n\} such that q_n \in U_n for all n. This subsequence is getting "closer and closer" to the point p_1, and since p_1 is in U_1, which is open, we know that eventually all the points in the subsequence will be in U_1. In other words, the subsequence \{q_n\} converges to p_1.

Now, we can repeat