upper semicontinuous definitions

Question

I'm learning about upper (lower) semicontinuous and saw $3$ defintions for this:
Definition 1:
The function $f: A\subset\mathbb{R}^n\to\mathbb{R}$ is called upper semicontinuous(resp.: lower semicontinuous) at $x_0$ if $\forall\epsilon>0,\exists\delta>0$ s.t: $f(x)-f(x_0)<\epsilon$ ( resp.: $f(x)-f(x_0)>\epsilon)$ $\forall x\in B(x_0,\delta)$; $f$ is upper semicontinuous (resp.: lower semicontinuous) on $A$ if $f$ is upper semicontinuous (resp.: lower semicontinuous) at every point in $A$.
Definition 2
a) $\forall a\in\mathbb{R}$, the set $\{x\in A: f(x)<a\}$ is open in $A$ (resp.: $\{x\in A: f(x)>a\}$ is open in $A$).
Definition 3
b) $\forall b\in\mathbb{R}$, the set $\{x\in A: f(x)\geq b\}$ is closed in $A$ (resp.: $\{x\in A: f(x)\leq b\}$ is closed in $A$).
I can see that def2 is equivalent to def3. But why def1 is equipvalent to def2 (or def3)? Could someone explain to help me understand better? Thanks in advance

Answer 1

While a bread-and-butter $\epsilon$-$\delta$ chase should be enough to prove this, here is a slightly more abstract way to think about this that will hopefully give some insight.

Let $\tau=\{(-\infty,a)\mid a\in \mathbb R\}\cup \{\emptyset,\mathbb R\}$. Note that $\tau$ is a topology on $\mathbb R$ (you can check the axioms by hand).

Then Definitions 2 and 3 simply state that $f$ is upper semicontinuous if and only if it is continuous as a map from $A$ (with the subspace topology induced from $\mathbb R^n$) into $(\mathbb R,\tau)$.

Now in general, a map $f$ between topological spaces is continuous at $x_0$ if and only if for every open neighborhood $V$ of $f(x_0)$, the pre-image $f^{-1}(V)$ is a neighborhood of $x_0$.*

The nontrivial open neighborhoods in $\tau$ that contain $f(x_0)$ are those of the form $V=(-\infty,f(x_0)+\epsilon)$ for some $\epsilon>0$, and $f^{-1}(V)$ is a neighborhood of $x_0\in A$ if and only if it contains some ball $B(x_0,\delta)$.**

Therefore $f$ is continuous at $x_0$ as a map into $(\mathbb R,\tau)$ if and only if for each $\epsilon>0$, we have $$f^{-1}((-\infty,f(x_0)+\epsilon))\supseteq B(x_0,\delta)$$ for some $\delta>0$.

That is, for all $x\in B(x_0,\delta)$, we have $x\in f^{-1}((-\infty,f(x_0)+\epsilon))$, or equivalently, $f(x)<f(x_0)+\epsilon$, which becomes $f(x)-f(x_0)<\epsilon$.

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(*We don't need to directly prove that the pre-image is open, and in the general case where we only have continuity at the point $x_0$, the pre-image may not be open.)

(**Note that here, and in your definition 1, $B(x_0,\delta)$ must refer to the ball in $A$, not in $\mathbb R^n$.)