If $f: I \to R$ is continuous and I is sequentially compact. Prove that if $\lim_{n\to\infty}f(x_n) = L $ then $\exists x \in I$ s.t $f(x) = L$ Verify
Yes, ordering everything a little: Since $(x_n)_n\subset I$, where $I$ is sequentially compact, then exists a subsequence $(x_{n_k})_k$ such that $x_{n_k}\to x$ for some $x\in I$. Now using the continuity of $f$, we have $f(x_{n_k})\to f(x)$. Since $f(x_n)$ is convergent and converges for $L$, then any convergent subsequence has to converge for the same limit and by the uniqueness of the limit we have that $f(x)=L$.