If every sequence of points of A admits a convergent subsequence, then every infinite subset of A has an accumulation point.

Question

I am studying for an exam and I came across the following question:

1. Having fixed an infinite subset A of $\mathbb{R}$, show that if every sequence of points of A admits a convergent subsequence, then every infinite subset of A has an accumulation point.

My idea to solve the problem is the following:

Suppose that exists a infinite subset $B \subset A$ such that $B' = \emptyset$. Then B is countable. Following that there exists a injective sequence

$$\phi : \mathbb{N} \longrightarrow B \subset A.$$

If exists $lim \phi = a$ then is a contradiction, because $a$ would be limit point of B

By assumption, some subsequence $(\phi_{n'})$ is convergent. Consider a crescent bijective function $f : \mathbb{N} \longrightarrow \mathbb{N}'$. Then the composition

$$\phi \circ f : \mathbb{N} \longrightarrow B$$

It is a convergent sequence of points of B. Again, we arrive at an absurdity, since the limit of this sequence would be an accumulation point of B.

I would like suggestions on how to get around this problem. I believe that my solution is incorrect, because I am creating a new function. My argument about being convergent is also very weak.

Answer 1

If I read the question correctly you are overthinking it.

Let $B$ be an infinite subset of $A$. Then $B$ has a countable subset you can list as a sequence. A convergent subsequence of that sequence accumulates at its limit.