I have done the proof that
$1)\ X$ is Frechet Compact iff $X$ is sequentially compact.
$2) \ X$ is sequentially compact iff $X$ is compact.
Thus we can conclude that $X$ is Frechet Compact iff $X$ is compact. Here Frechet Compact implies that $X$ is Frechet Compact if every infinite subset of $X$ has a limit point in $X$.
But I am trying to find a direct proof of $X$ is Frechet Compact iff $X$ is compact.Please Help.
Thank You!
I don’t think that there is a truly direct proof of the hard direction, but we don’t need to go through sequential compactness. I’ll outline the argument, leaving the details for you to try.
Suppose that the metric space $\langle X,d\rangle$ is Fréchet compact.
Show that for each $n\in\Bbb Z^+$ there is a finite $F_n\subseteq X$ such that $\left\{B\left(x,\frac1n\right):x\in F_n\right\}$ covers $X$. HINT: If not, use Zorn’s lemma to get an infinite set without a limit point.
Show that $D=\bigcup_{n\in\Bbb Z^+}F_n$ is a countable dense subset of $X$.
Show that $\left\{B\left(x,\frac1n\right):x\in D\text{ and }n\in\Bbb Z^+\right\}$ is a countable base for $X$.
Conclude that $X$ is Lindelöf.
Now let $\mathscr{U}$ be any open cover of $X$. $X$ is Lindelöf, so $\mathscr{U}$ has a countable subcover, and we might as well simply assume from the start that $\mathscr{U}$ is countable. If $\mathscr{U}$ is finite, we’re done, so suppose that $\mathscr{U}=\{U_n:n\in\Bbb N\}$. For $n\in\Bbb N$ let $V_n=\bigcup_{k\le n}U_k$; clearly $V_n\subseteq V_{n+1}$ for each $n\in\Bbb N$, and $\bigcup_{n\in\Bbb N}V_n=X$. If $V_n=X$ for some $n\in\Bbb N$, then $\{U_k\in\mathscr{U}:k\le n\}$ is a finite subfamily of $\mathscr{U}$ covering $X$.
If not, there is a strictly increasing sequence $\langle n_k:k\in\Bbb N\rangle$ in $\Bbb N$ such that $V_{n_k}\subsetneqq V_{n_{k+1}}$ for each $k\in\Bbb N$. Thus, for each $k\in\Bbb N$ we can choose a point $x_k\in V_{n_{k+1}}\setminus V_{n_k}$.