$U=\{x\in \mathbb{R^n} :g_i(x)\leq 0 : \text{for} \quad i=1,...,m \}$ is closed

Question

Let $U=\{x\in \mathbb{R^n} :g_i(x)\leq 0 : \forall i=1,...,m \}$ $\subset$ $\mathbb{R^n}$.

with $(g_i)_{1\leq i \leq n}$ are convex functions from $\mathbb{R^n}$ in $\mathbb{R}$.

we want to prove that $U$ is a closed set.

Answer 1

Any convex function from $\mathbb R^{n}$ to $\mathbb R$ is continuous. It follows that $U$ is the intersection of a $m$ closed sets, hence closed.

Answer 2

I think we have $g=(g_1,...,g_m): \mathbb R^n \to \mathbb R^m$ and $U = \{x \in \mathbb{R}^n : g_i(x) \leq 0, \forall i = 1, \dots, m\}$ with each $g_i$ convex.

Let $U_i := \{x \in \mathbb{R}^n : g_i(x) \leq 0\}$.

Since $g_i$ is continuous, we have that $U_i$ is closed. Hence $U= \bigcap_{i=1}^n U_i$ is closed.