Let $X=\left\{\left(x_1, x_2, ..., x_n\right) \in \mathbb{R}^n \vert x_i \geqslant 0, \forall i \mbox{ and }x_1+x_2+...+x_n=1\right\}$. Consider $f: X\rightarrow \mathbb{R}$. There is a proof in a paper claiming if $f$ is bounded, then the hypograph of $f$ is path connected. It does not elaborate on it so I was wondering how one can prove it. I tried it but got confused. Path connectedness involves continuity, but $f$ is not necessarily continuous.
Edit: The hypograph of $f$ is $H=\left\{\left(x_1, x_2, ..., x_n, y\right)\in X \times \mathbb{R} \vert y \leqslant f\left(x_1, x_2, ..., x_n\right)\right\}$. Consider $\left(x; y\right), \left(x';y'\right)\in H$. A path is a continuous function $p: \left[0,1\right]\rightarrow H$ such that $p\left(0\right)=\left(x;y\right)$ and $p\left(1\right)=\left(x';y'\right)$.
My attempt: I think I can focus on the case where the projection of the range of $p$ on $X$ is the segment between $x$ and $x'$. I don't know any theorem in the form of "there exists a continuous function such that...". So maybe I should directly construct a path. But I don't know how to solidly describe such a construction, not to mention to ensure its continuity given that $f$ is not necessarily continuous.
More generally, suppose $X$ is any path-connected space and $f:X\to\mathbb{R}$ is a bounded-below function. Then the hypograph $H$ of $f$ is connected. Indeed, let $a\in\mathbb{R}$ be such that $f(x)\geq a$ for all $x\in X$ and suppose $(x,r),(y,s)\in H$. To get a path from $(x,r)$ to $(y,s)$ in $H$, take a path from $(x,r)$ to $(x,a)$ where only the second coordinate varies (which is possible since $f(x)\geq a$ so $(x,t)\in H$ for all $t$ between $r$ and $a$), then a path from $(x,a)$ to $(y,a)$ where only the first coordinate varies (which is possible since $X$ is path-connected), and finally a path from $(y,a)$ to $(y,s)$ where only the second coordinate varies.