Trouble Understanding: Lower Semi-Continuity Subsequence Proof

I'm reading the following proof, but I can't seem to follow one line (the last line listed). The proof is as follows:

Assume by contradiction that $$f$$ is not lower semicontinuous, meaning that there exists $$x^{*} \in E$$ and $$\{x_n\} \subseteq E$$ such that $$x_n \rightarrow x^{*}$$ and $$\liminf_{n \rightarrow \infty} f(x_n). Take $$\alpha$$ that satisfies:$$\liminf_{n \rightarrow \infty} f(x_n)<\alpha

Then, there exists a subsequence $$\{x_{n_k}\}_{k \geq 1}\rightarrow x^{*}$$ such that $$f(x_{n_k}) \leq \alpha$$ for all $$k \geq 1$$.

I'm having trouble seeing why does there exist a subsequence $$\{x_{n_k}\}$$ such that $$f(x_{n_k}) \leq \alpha$$. I revised my undergraduate real analysis notes but I couldn't figure it out. I'm guessing it is related to $$\liminf_{n \rightarrow \infty}$$ but I can't seem to figure it out. Where does it come from? Thanks.

Suppose there is no such subsequence $$\{x_{n_k}\}$$; then we know that $$f(x_n) \gt \alpha$$ for every $$n \in {\mathbb N}$$. But then $$\liminf_{n\rightarrow \infty} f(x_n)$$ cannot be less than $$\alpha$$, which contradicts how we chose $$\alpha$$.
The key idea here is that the strict inequality $$\liminf f(x_n) \lt f(x^*)$$ guarantees us some number lying between our two values (a bit like finding a separating hyperplane), and then the limiting process guarantees us that we must cross the hyperplane.
• This is a really nice answer thanks. If I understand correctly if $f(x_n)>\alpha$ for every $n\in \mathbb{N}$ then this implies $\liminf_{n \rightarrow \infty} f(x_n)>\alpha$ which is a contradiction. Is this correct? Apr 28 '20 at 14:02
• @yessssir It implies $\liminf_{n\rightarrow \infty} f(x_n) \geq \alpha$ but yes (think about $1/n$ for $n\in {\mathbb N}$ which is $\gt 0$ but the $\liminf$ is $0$) Apr 28 '20 at 14:04