Let $f:\mathbb{R} \mapsto \mathbb{R}$ be a lower semi-continuous function. Prove or disprove that if $X \subset \mathbb{R}$ has a lower bound, then $f(X)$ also has a lower bound.
My attempt:
So far, every example that I could think of a lower semi-continuous function that might be a counter-example ends up satisfying this property, e.g., $f(x) = \frac{1}{|x|},x\neq0,f(0)=0$, so I've been trying to prove that it is true instead of looking for counter-examples.
If I assume that $\inf{f(X)}$ is finite, then the problem is easy: just take a sequence in $f(X)$ converging to $\inf{f(X)}$ and the properties of lower semi-continuity will guarantee the statement. However I can't make such an assumption. I've tried considering a sequence $(x_n) \subset X, x_n \rightarrow \inf{X}$ and use that $\text{lim inf}_{n\rightarrow \infty}f(x_n) \geq f(\inf{X})$, but I wasn't able to conclude anything from this.
Any hints or suggestions would be appreciated.
This proposition is false. If $f$ is $x\mapsto-x$ (or any continuous function such that $\liminf_{x\to\infty}f(x)=-\infty$), then $f$ is lower semi-continuous but $f((0,\infty))=(-\infty,0)$ does not have a lower bound.