# Math Help - Closed subset of sequentially compact set is sequentially compact.

1. ## Closed subset of sequentially compact set is sequentially compact.

Let B be a subset of Y, and let B be sequentially compact. Let D be a closed subset of B. Then D is sequentially compact.

Can anyone help me out and give me a general direction to go with proving this?

2. ## Re: Closed subset of sequentially compact set is sequentially compact.

Let $(x_n)\subset D$ be an infinite sequence, since $B$ is sequentially compact there is a subsequence (I'll call it the same) and $x\in B$ such that $x_n \to x$, but $D$ is closed so...