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

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.

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.
• It's easy to see from the definition of convergence that the limit of a sequence is an accumulation point of the sequence. For every $\epsilon$ there will be (infinitely many) elements from the sequence withing $\epsilon$ of the limit. That's what it means to be an accumulation point. Commented Aug 14, 2023 at 18:52