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