# A subspace $Y$ of a complete metric space $X$ is closed if it is complete.

For this problem, let there is a sequence $\{y_n\}$ which is complete in $Y$ such that this sequence has finitely many distinct points and this sequence approaches to say $y$. In this case $y$ is repeated infinitely many times. In this scenario, there may exist open ball $B(y;r)$ which only contains $y$ and so $y$ can't be the accumulation point of this sequence. So however $Y$ is complete, it is not closed.

I am not able to understand the mistake.

-
What does this have to do with descriptive set theory? –  Chris Eagle Aug 25 '12 at 17:02

Completeness is not a property of sequences; I’m going to assume that you mean that $\langle y_n:n\in\Bbb N\rangle$ is a Cauchy sequence. If it has only finitely many distinct terms and is Cauchy, then there are a $y\in Y$ and an $n_0\in\Bbb N$ such that $y_n=y$ for all $n\ge n_0$. (You should try to prove this.) In other words, the sequence is eventually constant at $y$. Then every open ball $B(y,r)$ contains $\{y_n:n\ge n_0\}$, so of course the sequence converges to $y$. In other words, $y$ is the limit of the sequence. Note that $y_{n_0},y_{n_0+1},y_{n_0+2},\dots$ are different terms of the sequence even though they all happen to be the same point. $\langle 1,1,1,\dots\rangle$ is an infinite sequence of real numbers even though it’s a constant sequence.