am working on a proof to present to a reading group I am in, that relies on the following proposition, which I am having trouble proving. Perhaps some of the brilliant minds that frequent this forum may find an opportunity to help.

Suppose that M(x) is an USC correspondence from S-->T, S in Rm, T compact in Rn.

I want to claim that for every x(n)-->x, there exists a sequence

y(n):y(n) is an element of M(x(n)), which converges.

Since T is compact, for ANY sequence y(n) st y(n) in M(x(n)), there exists a SUBsequence which converges, but I need an element y(n) in
M(x(n)) for EVERY x(n).

In case the reader is wondering, I am trying to prove that if f is continuous and M is usc, then the correspondence f(x,M(x)) is also usc, which in turn I am using to prove the Berge "theorem of the maximum." Any ideas would be much appreciated

> am working on a proof to present to a reading group I am
>in, that relies on the following proposition, which I am
>having trouble proving. Perhaps some of the brilliant minds
>that frequent this forum may find an opportunity to help.
>
>Suppose that M(x) is an USC correspondence from S-->T, S in
>Rm, T compact in Rn.
>
>I want to claim that for every x(n)-->x, there exists a
>sequence
>
>y(n):y(n) is an element of M(x(n)), which converges.
>
>Since T is compact, for ANY sequence y(n) st y(n) in
>M(x(n)), there exists a SUBsequence which converges, but I
>need an element y(n) in
>M(x(n)) for EVERY x(n).
>
>In case the reader is wondering, I am trying to prove that
>if f is continuous and M is usc, then the correspondence
>f(x,M(x)) is also usc, which in turn I am using to prove
>the Berge "theorem of the maximum." Any ideas would be much
>appreciated

I think the Closed Graph theorem is of relevance here. For S and T compact, M is USC iff the graph of M is closed.

Take M(x) = sin(pi/x) on some [-N, N], x != 0, and M(0) = [-1,1].

Since the graph of M is closed, M is USC. Now, consider,

xn = 1/n.

{M(xn)} consists of infinitely many 1's, 0's, and -1's. It clearly contains convergent subsequences, but does not constitute a convergent sequence itself.