I've been trying to get my head around this for a bit but I am not sure where to start. Here is the question:
Prove the following theorem:
Let $Y$ be a closed and convex subset of $\mathbb{R}^n$ such that $0 \in Y$ and $ Y \cap \mathbb{R}^n_{+} = \{0\}$. For each $i = 1, 2.... n$ define a correspondence $Y_i:\mathbb{R}^{n-1} \mapsto \mathbb{R}^n$ as: $$ Y_i(z) = \{y \in Y: y_{-i} = z \} $$
Then for each $ i = 1, 2, ... n$ and each $ z \in R^{n-1}$, the set $Y_i(z)$ is closed and there exists $ M \in \mathbb{R}$ such that for all $y \in Y_i(z)$, $y_i < M$
I am kinda struggling to see why the correspondence $Y_i(z)$ would be closed.
I've tried this approach:
Assume $Y_i(z)$ isn't closed
Then there exists ${y_n} \in Y_i(z)$ such that $y_n \mapsto y'$ and $y' \notin Y_i(z)$. But we know $y' \in Y$. And I honestly don't know where to go from here. Any help or clue would be great. I am trying to learn how to get comfortable with real analysis and related problems more, so any thoughts on how to even think about this would be great.
Here is my attempt at an answer. It isn't fully formalized and I am not sure if I am correct.
If $Y_i(z)$ is closed for a particular $z$ and $i$ then we know a couple of things:
$\exists \ \{\hat{y}_k\} \in Y_i(z)$
such that
$ \hat{y}_k \rightarrow \hat{y} $
where $\hat{y} \in Y_i(z)$
It's also the case that all of these variables exist in $Y$
It also has to be that $\hat{y}_{-ik} = z$. That's how it got selected in $Y_i(z)$
For the entire sequence all the elements of each vector except the $i^{th}$ element are constantly just $z$.
So the only element we actually have to check for convergence is the $i^{th}$ element.
Let $\hat{y}_{ik} \rightarrow y^*_i$ and $y^*_i \in y^*$
Let $y^*_{-i} = z$. This is so the whole vector $y^*$ is the limit of $\hat{y}_k$.
Since $y^*_{-i} = z$ we know $y^* \in Y_i(z)$. Therefore the limit of the sequence {\hat{y}_k } exists within $Y_i(z)$. This is for any arbitrary $i$ and $z$. Therefore $Y_i(z)$ is closed for all $i$ and $z$.
For the second bit, I believe it's trivial from here:
$y_{ik} \rightarrow y^*_i$, which means that $ \forall \epsilon > 0 \ \exists \ N \in \mathbb{N}$ such that $ \forall n \in N \ d(y^*_i, y_{ik}) < \epsilon$.
That means we can find a big enough $\epsilon$ such that $N = 0$. That is, the whole sequence is less than $\epsilon$ distance away from $y^*_i$. Therefore there will exist some $M \in \mathbb{R}$ such that $y_ik < M$ for all $k$.