Lower semicontinuity of a multi-valued map $F:X\to 2^Y$ in term of net

Question

Let $X,Y$ be two Hausdorff spaces and $F:X\to 2^Y$ be a multi-valued mapping. We says that $F$ is lower semicontinuous at $x_0\in X$ if for each $y_0\in F(x_0)$ and any neighborhood $U\in \mathcal N(y_0)$, there exist a neighborhood $V$ of $x_0$ such that $$ F(x) \cap U\neq \emptyset \quad\text{for all}\quad x \in V. $$

In Aubin and Cellina's Differential Inclusion, it is claimed that the property is equivalent to

given any generalized sequence $x_\mu \to x_0$ and any $y_0\in F(x_0)$, there exists a generalized sequence $y_\mu \in F(x_\mu)$ such that $y_\mu\to y_0$.

I assume that they meant net when they said generalized sequence.

I don't find this claim to be trivial at all. In fact, I found a proof of this from a different source that I believe to be invalid.

Let $(x_\mu)_{\mu\in \mathcal I}$ be a net converging to $x_0$, the aforementioned proof construct a net $y:D\to Y$ from a directed set $D\subset \mathcal I \times \mathcal N(y_0)$ defined by $$ D:=\{(\mu,V) : F(x_\mu)\cap V \neq\emptyset \} $$ by choosing $$ y_{\mu,V} \in F(x_\mu)\cap V. $$ It is not hard to verify that $(y_{\mu,V})_{(\mu,V)\to D}$ converges to $y_0$. However, their last step of finding a map $\sigma:\mathcal I \to D$ to compose with $y$ seems invalid, since they let $\sigma$ be a arbitrary section of $\pi:D\to \mathcal I$, where $\pi$ is the projection $\pi(\mu,V)=\mu$.

However, I have looked around and found an old paper by Frolík, Concerning topological convergence of sets, from 1960. It contains a statement that implies a weaker result (but seems more reasonable to me) that could be paraphrased as follows.

Let $(M_\mu)_{\mu\in \mathcal I}$ be a net in $2^Y$. Then $y_0$ is in $\liminf M_\mu$ if and only if there exists a net $\pi:\mathcal I'\to \mathcal I$ which is residual in $\mathcal I$ and points $y_\lambda \in M_{\pi(\lambda)}$ such that $(y_\lambda)_{\lambda\in\mathcal I'} \to y_0$.

The proof is almost identical to the previous one but he stopped by letting $\mathcal I'=D$, defined as in the above proof, and concluded the theorem without having to invoke a section $\sigma$ of $\pi$.

Question: Is the claim by Aubin and Cellina true? If it is, then where can I find a valid proof of it?

Answer 1

You're probably right about the statement of the theorem. Indeed, many textbooks (especially older ones) call nets "generali(s/z)ed sequences" I think it should say:

Given a net $x: \mathcal{I} \to X$ such that $x \to x_0$ in $X$, there is a directed set $\mathcal{D}$ and a net $y:\mathcal{D} \to Y$ such that $y \to y_0$ and with a "connecting function" $\pi: \mathcal{D} \to \mathcal{I}$ as in a Kelley style subnet, such that
$y(d) \in F(x(\pi(d))$ for all $d \in \mathcal{D}$.

By the Kelly style connecting function $\pi$ I mean the condition that $$\forall d \in \mathcal{D} \exists \mu_0 \in \mathcal{I}: (\forall \mu \ge \mu_0: \pi(\mu) \ge d)$$

which is weaker than what some authors (like Willard use), namely monotonous with cofinal image. In fact the one in your sketched proof is of that stronger form. Note that also Frolik uses this connecting map idea. It seems the natural way to go.

So we need a larger index set for the net converging to $y$ but of course the same $\pi$ at least allows us to use the same index set for a net converging to $x_0$ and one converging to $y_0$, as $x' = x \circ\pi$ is a subnet of $x$ so also converges to $x_0$.

You cannot use an arbitrary "lifting" of $\pi$ to achieve this: the idea of the construction of $\mathcal{D}$ as such a subset of a product is also used in showing that a net with an adherent point $p$ has a subnet converging to that point (using reversely ordered neighbourhoods of $p$, as we do here with $y_0$). But we cannot always garantuee the same index set as we start with, as any compact non-sequentially compact space shows: there we have sequences with convergent subnets, but without convergent subsequences. So there we're sure we cannot lift the projection, and the great similarity between these situations seems to warrant (to me at least) that such a step (some one-sided inverse of $\pi$ to get back to $\mathcal{I}$) cannot always be made.