Given the set
$V(q)=\{(x_1,x_2) \in \mathbb R^2:ax_1≥\log y \text{ and } bx_2≥\log y\}$
with $a$, $b$, and $y$ strictly positive, I have to show that $V$ is closed and convex.
My idea for convexity is to show that:
$z=t\cdot ax_1+(1-t)\cdot bx_2≥\log y$
with $t∈[0,1]$ (or $t∈(0,1)$?) so that $z∈V(q)$.
I do not have ideas at the moment on how to show that is closed and convex, could you help me?
EDIT: in general, given a vector of inputs $(x_1,…,x_{L-1})$, only one output $y$, $V$ can be defined as
$V(q)=(x∈ℝ_+^{L-1}:(y,-x)∈Y)$
where $Y$ is the set of all combinations of inputs and outputs. $V$ can take various mathematical forms, which one is that reported in the text of my question.
Best Answer
Recall that if $f: X \rightarrow Y$ is a continuous map between, say, metric spaces, then the pre-image $f^{-1}(C)$ of a closed set $C$ is closed. Applying this, the set $V$ is closed because it is the intersection of pre-images of the closed set $[\log y, \infty)$ under the continuous functions $(x_1, x_2) \mapsto ax_1$ and $(x_1,x_2) \mapsto bx_2$. As far as convexity goes, your approach is almost correct. Let $$t(x_1, x_2) + (1-t)(x_1', x_2') = (tx_1 + (1-t)x_1', tx_2 +(1-t)x_2)$$ be an element of the line between $(x_1, x_2)$ and $(x_1', x_2')$, both of which are assumed to lie in $V$. Compute for the first cooredinate, using $x_1,x_1' \geq \log y$, $$a (tx_1+(1-t)x_1') = tax_1 + (1-t)ax_1' \geq t\log y + (1-t) \log y = \log y.$$ Similarly, for the second coordinate, $$b(tx_2 + (1-t)x_2') \geq \log y.$$