I have a problem with the definitions of upper hemicontinuity. In particular I found a picture that makes me wonder, not only about my understanding of this concept, but of the concept and application of sequences in general.
Just as a side note, I am self-taught, thus I am sorry if my question will sound really either obvious or naive.
The problem is the following. Consider the following picture. This correspondence is LHC, but it is not UHC. I can see from the picture itself why it is LHC, but it is not clear to me why it is not UHC.
[Just as a side note, in order to check if it is UHC, the image of the correspondence has to be compact, and the correspondence $f: X \rightrightarrows Y$ has to have a closed graph. This means that, if for any $x^m \in A$, $x^m \to x$, and $y^m \in f (x^m)$, $y^m \to y$, then $y \in f(x)$.]
Here there is my guess. If we focus on the correspondence whose domain is $(2, \infty)$, we can get a sequence $y^m$ that converges to $y$ from the right, where $y \neq 2$. For example, let’s say that the upper bound of the correspondence on $(2, \infty)$ is $2 \frac{2}{3}$, then (from the right) $y^m \to 2 \frac{2}{3}$ , but $2 \frac{2}{3} \notin f(\frac{1}{2})$.
As you can guess, sequences make me feel slightly uncomfortable. It is like I am not sure if it is “allowed” to move from the right to the left, not only when we play around with a definition, but in general. [Indeed, this sounds naive]
Thus, does this make sense?
Thanks a lot in advance as always for any help or feedback.
Best Answer
Unlike you, I can't put a picture. So my explanation will be verbal. I am going to use topological definiton of uhc, that is, if $f$ is uhc at $x$ than for arbitrary open neighborhood $U$ of $f(x)$ we can find an open neighborhood $V$ of $x$ such that for $x^{\prime}\in V$ we have $f(x^{\prime})\subset U$. This definition implies what you gave above (closed-graph theorem). So consider the point $1/2$ and its image $2$. No matter how small $\epsilon$,or $U$ is there is no $\delta$, or $V$ for $1/2$ that satisfies definition, because every $V$ contains points higher than $1/2$ and their image is not subset of $U$.