A subset $E$ of $X$ is said to be sequentially compact if and only if every sequences $x_n \in E$ has a convergent subsequence whose limit belongs to $E$.
Prove that every sequentially set is closed and bounded.
Attempt: Suppose $ E$ is a subset of $ X$ is sequentially compact set.
Let $x_n \in E$ be a sequence in $E$.
So there is subsequence ${x_{n_{k}}}$ of ${x_n}$ which is convergent and its limit belongs to $E$. Thus by a theorem $E$ is closed if and only if the limit of every convergent sequence $x_n \in E$ satisfies limit$_{k → \infty} x_n \in E$. So E is closed.
Then we need to show E is bounded.
By contradiction suppose E is not bounded.
So let $x_n \in E$ and $a \in E$ such that $d(x_n , a) > n$ for all $n \in N$. Since $E$ is a sequentially compact space and $x_n \in E$ is a sequence so there is a subsequence ${x_{n_{k}}}$ of ${x_n}$ in E which is convergent and converges to a point in $E$.
Then $d({x_{n_{k}}}, a) \leq d({x_{n_{k}}}, b) + d(b,a) < \epsilon + d(b,a). $ by triangel inequality.
But $d({x_{n_{k}}}, a) > n_k$ for all $k \geq N$ as $k → \infty$.
So ${x_{n_{k}}}$ diverges which is a contradiction. So $E$ is bounded.
Is this correct? can anyone please help me? Any feedback would be really appreciate it.
Thank you.
Best Answer
For the bounded part I think it's brilliant and ingenious! (You still can show that the function $d(a,x_{n})$ is continuous, but it's not important). As for closure, your sequence $x_{n}$ must be convergent to be able to conclude. eg: Let $x_{n}$ be a sequence in E which converges in X. Then $x_{n}$ has a convergent subsequence, and by uniqueness of limit, your conclusion is perfect! (Equivalenty, we can show that Cl(E)=E. For, let $x \in$ Cl(E). Then there exists a sequence $x_{n}$ in E converging to x, and then it's the same argument as above!