Equality of definitions of Lower Semicontinuity I would like to show the equality (i.e. iff) of the following definitions of lower semicontinuity without the use of liminf.
Definition: Lower Semicontinuous
Let $(X, \|\cdot\|)$ be a normed vector space with a topology $\tau(X)$ induced by said norm. A function $f:X\rightarrow \mathbb{R}$ is lower semicontinuous if

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*$\forall(x_{0}\in X, \epsilon > 0)~~\exists ~ \delta>0~$ s.t. $\forall x\in X~~||x_{0}-x||<\delta \implies f(x_{0})-f(x) < \epsilon$

*$\{x\in X:f(x) \leq \alpha \}$ is closed in X,$~$ i.e. $\{x\in X:f(x) \leq \alpha \}^{~c} \in\tau(X)$,  $~\forall\alpha\in \mathbb{R}$
$~$
Here are similar questions and useful links:

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*closed epigraphs equivalence

*Show the iff statement of lower semicontinuous

*https://www.encyclopediaofmath.org/index.php/Semicontinuous_function

*https://proofwiki.org/wiki/Characterization_of_Lower_Semicontinuity
 A: Suppose the condition 1 holds true, and let $\alpha \in \mathbb R$. If $x_0\in f^{-1}((\alpha, +\infty))$, then let $\varepsilon =f(x_0)-\alpha$. By hypothesis, there is some $\delta>0$ such that $f(x)>f(x_0)-\varepsilon=\alpha$ whenever $\lVert x-x_0\rVert<\delta$. Equivalently, the ball of radius $\delta>0$ centered at $x_0$ lies in $f^{-1}((\alpha,+\infty))$. Since this is true for all $x_0\in f^{-1}((\alpha,+\infty))$, we conclude that $f^{-1}(\alpha,+\infty)$ is open, that is, $f^{-1}((-\infty, \alpha])$ is closed. Thus the condition 2 holds true.
Conversely, suppose that the condition 2 holds true, and let $x_0\in X$. Given $\varepsilon>0$, the set $f^{-1}((-\infty,f(x_0)-\varepsilon])$ is closed, and thus its complement $f^{-1}((f(x_0)-\varepsilon,\infty))$ is open. Because $x_0\in f^{-1}((f(x_0)-\varepsilon,\infty))$, this means that there is some $\delta>0$ such that the ball of radius $\delta$ centered at $x_0$ lies in $f^{-1}((f(x_0)-\varepsilon,\infty))$. This $\delta $ satisfies the requirement in the condition 1.

The proof is very, very similar to the proof that continuity of a function is equivalent to continuity at all points in the domain. In fact, $X$ need not be a normed vector space, but any metric space.
