Let $g_i(x):\mathbb{R}^{n} \to \mathbb{R}$, for $i=1,\ldots,n$, be continuous convex functions. Define $g_{\rm max}$ as $g_{\rm max}(x) \triangleq \mbox{max}_{i=1,\ldots,n}\{g_i(x)\}$. Define also the following feasible sets: $$\mathcal{F}_1 \triangleq \{x \in \mathbb{R}^n: g_i(x)\leq0 ~~\forall i \}$$ and $$\mathcal{F}_2 \triangleq \{x \in \mathbb{R}^n: g_{\rm max}(x)\leq0 \}$$
Which statements are true (if any)? Why?
1) $\mathcal{F}_1 \subseteq \mathcal{F}_2$
2) $\mathcal{F}_1 \supseteq \mathcal{F}_2$
3) Statementes 1 and 2, i.e., $\mathcal{F}_1 = \mathcal{F}_2$
I think all statements are true as $$x^* \in \mathcal{F_2} \Rightarrow g_{max}(x^*) \leq 0 $$ but, $0 \geq g_{max}(x^*) \geq g_i(x^*)$ for all i. Then, $\mathcal{F}_2 \subseteq \mathcal{F}_1$.
On the other hand, take $y \in \mathcal{F}_1$, then $g_i(y) \leq 0$ for all $i$ and as $i = 1,...,n$ is finite then $g_{max}(y) = g_j(y)$ for some $j$, then $g_{max}(y) \leq 0$. Then, $\mathcal{F}_1 \subseteq \mathcal{F}_2$.
Were we conclude that, $\mathcal{F}_2 = \mathcal{F}_1$.