# Proof verification: lower semi continuous at x implies that the lim inf of every convergent sequence is bigger than f(x)

Show that f is lower semi-continuous at x if, and only if, $$\liminf f(x_n) \geq f(x).$$ for any sequence $(x_n)$ converging to $x$.

Proof (only if)*:

Assume that $f$ is lower semi-continuous at $x$. Let $(x_n)$ be an arbitrary sequence converging to $x$. Fix an arbitrary $\varepsilon > 0$. Since $(x_n)$ converges to $x$, $\exists N \in \mathbb{N}$ s.t. $\forall n\geq\mathbb{N}, \ \ |x_n - x| < \delta$

We have that by lower semi-continuity, this implies that (for every $n$ bigger than $N$) $f(x_n) - f(x) > -\varepsilon$. Then it is the case that $$\liminf f(x_n) - f(x) > -\varepsilon.$$ Since we can take $\varepsilon$ to be as small as we want, we have that $$\liminf f(x_n) \geq f(x).$$

Since if we assume by contradiction that $f(x) >$ lim inf $f(x_n)$, we get the existence of an $\varepsilon$ and an $\delta$ such that lim inf $f(x_n) - f(x) < -\varepsilon$.

Does this proof hold?

If not, which are the mistakes and the alternatives?

Thank you

*Proof of if comes by contradiction without much worries.

Now, after you established that $\liminf_nf(x_n)\geq f(x)$, you keep going with "since if we assume by contradiction..." I don't see how that contributes anything to the proof.