Continuity of correspondence - Hemi-continuity I’ve been going through the various definitions of upper and lower hemicontinuity, viz. the neighborhood definition, the sequential characterization and the definition in terms of the upper and lower inverse being open sets. However, I think I’m not quite clear on the definitions as I’ve been looking at several cases and I’m unclear about how to reconcile two seemingly similar cases.
Case 1: This is from a YouTube lecture/video from Columbia. He describes why the following is not upper semi continuous, and this makes sense to me:

Case 2: This is from the book Real Analysis with Economic Applications by Efe A. Ok, specifically from page 288:

To me, these two cases look somewhat similar. However, it’s not clear to me why in Case 2, the second case, $\Gamma_2$ is upper semicontinuous at $x_2$. It seems to me that if you take a neighborhood around $x_2$ that we can find points that are not in the union of the neighborhoods around the image under $\Gamma$ there, similar to what we see in the figure in Case1.
What am I missing?
(Explaining using just one of the definitions would be great, but iff you’re able to give me the clues for all 3 definitions that would help my understanding).
 A: I will attempt to answer using the sequential model, but first let
me fix some of the points from the youtube lecture's correspondence. Let us call the top of the base of the triangle as $V_{1}$. The point mid way through the base is $V_2$ (s.t. $V_{1}$ and $V_{2}$ is the solid lines). $V_{3}$ is the point at the bottom (s.t. $V_{1}$,$V_{2}$ and $V_{3}$ all lie on the base of that triangle). The shaded area is all part of the correspondence output. Note : $\Gamma(x)$ = $[V_{1},V_{2}]$ $\cup$ $V_{3}$.
Recall the definition of UHC (I give a rough definition) , if you pick a sequence $x_{n}$ converging to $x_{3}$ and if you pick a sequence of $y_{n}$ $\in$ $\Gamma_{n}(x_{n})$ converging to $y$, then it MUST be the case that $y$ $\in$ $\Gamma(x)$. Let us go about creating such a sequence to disprove UHC,remember you only need one.
Consider the sequence $y_{n}$ = $[V_{1}-\frac{1}{n},V_{3}-\frac{1}{n}]$ and $x_{n} = \{x_{3}+\frac{1}{n}\}$. Now, $[V_{1}-\frac{1}{n},V_{3}-\frac{1}{n}] \rightarrow [V_{1},V_{3}]$ and  $\{x_{3}+\frac{1}{n}\} \rightarrow x_{3}$.  Notice that for the closed intervals
$[V_{1}-\frac{1}{n},V_{3}-\frac{1}{n}]$ lie in the shaded area, therefore $[V_{1}-\frac{1}{n},V_{3}-\frac{1}{n}]$ $\in$ $\Gamma(x_{n})$ . We need to establish that the limit point $[V_{1},V_{3}]$ $\in$ $\Gamma(x_{3})$. Now notice again that, $\Gamma(x)$ = $[V_{1},V_{2}]$ $\cup$ $V_{3}$. Since the limit is the closed interval plus single point $V_{3}$, the limit of the sequence we chose has a whole extra bit $(V2,V3)$ that is not there in $\Gamma(x_{3})$ but if you just go a tiny bit to the right to some point $x_{3}+\epsilon$ the interval $[V_{1}-\frac{1}{n},V_{3}-\frac{1}{n}]$ $\in$ $\Gamma(x_{3}+\epsilon)$ (for some 'n') , it just does not lie in the output of the correspondence AT the point $x_{3}$.
Two things :
I have done this from the right side of $x_{3}$, I believe the left side is trivial (it seems to me that you may not know that the shaded area is part of the correspondence).
Adding in the set theoretic definition as well, recall, for any $V$ s.t. $\Gamma(x_{3})$ $\in$ V $\subseteq$ $\Gamma(x)$ (we are just choosing a subset of your y-axis which contains the image of the correspondence at $x_{3}$, with as little notation as possible), we should be able to find a set (in this case a neighbourhood since you are on the real line) around $x_{3}$ such that all the images of the the neighbourhood lie within the V you chose. So let's choose a $V$
Step 1: Choose $V$  = $\Gamma(x_{1})$, note that it satisfies all the conditions required above. Note that it is just a singleton set.
Step 2: Now try to construct a neighbourhood around $x_{3}$ on the x-axis such that the image of all these points is contained in $V$. YOU CAN'T! The reason being that any image around $x_{3}$ contains an output of the correspondence which are larger sets, containing much more than a single point as soon as you go a little $\epsilon$ distance to the right of $x_{3}$.Since the image of these points is not a subset of $\Gamma(x_{3})$, this function is not UHC at $x_{3}$.
You can try creating an example along similar lines for Efe's illustration as well, if you can it's UHC. But you cannot! Happy to give an example of that as well, just comment.
