Is the space of all convergent sequences compact in the space of all bounded sequences($l^{\infty}$)?
Argument: I think the answer is in yes because in particular if we consider a sequence of all complex numbers numbers then that sequence will contain both bounded and unbounded sequences,Sice all convergent sequences are bounded this means that convergent sequences will become a part of bounded sequences. Now, we know that $\mathbb C$ is a Hausdorff space and set of all convergent sequences is closed.Now we can get support from the result Closed subspace of a Hausdorff space is compact, making space of all convergent sequences compact in $l^{\infty}$.
Is it correct??
Please give your opinions...
First of all, the result " a closed subspace of a Hausdorff space is compact" is clearly wrong: take any Hausdorff non compact space $X$, then $X$ is closed in $X$, but not compact. There are plenty of examples of such spaces : $\mathbb{R, C, Q}$ probably being the most basic you should think of.
Now the result you ask about is also wrong. Indeed, $\mathbb{C}$ clearly embeds into this space as a closed subspace (the embedding being given by $z\mapsto (z)_{n\in \mathbb{N}}$: each of these is indeed convergent, and with the sup-norm this is clearly an isometric embedding whose image is closed); but $\mathbb{C}$ is not compact, so the space can't be.