Is this set in $\mathbb{R}^d$ closed?

Question

Let $X$ be a convex set and for each $ i \in \{1,2,\dots,d \} $, let $f_{i}: X \rightarrow \mathbb{R} $ be convex functions where $ C_{i} := \{ f_{i}(x) : x \in X \} = [a_{i}, b_{i}]$ - closed interval in $\mathbb{R}$.

Does that mean the set $ C:= \{ \left(f_{1}(x), f_{2}(x),\dots, f_{d}(x)\right) : x \in X \} \subseteq \mathbb{R}^{d} $ is closed?

Answer 1

This set in $\mathbb R^2$ is convex: $$ X = \{(x,y) : x^2+y^2 < 1\} \cup \{(1,0),(0,1),(-1,0),(0,-1)\} $$ These two maps are linear, hence convex: $$ f_1(x,y) = x,\qquad f_2(x,y) = y $$ Of course $$ f_1(X) = [-1,1],\qquad f_2(X) = [-1,1]. $$ But $C = X$ so it is not closed.