Proving that $f$ has a minimum value Let $f$ be a real-valued function(not necessarily continuous) on $\mathbb R^n$ with the property that for any $x\in \mathbb R^n $ and $\epsilon>0$, there is $\delta>0$ such that $f(y)>f(x)-\epsilon$ for all $y\in \mathbb R^n$ with $||y-x||<\delta$. Show that if $\lim_{||x||\to\infty}f(x)=1\ge f(c)$ for some $c \in \mathbb R^n$, then $f$ has a minimum value on $\mathbb R^n$.
I have no idea how to deal with this question. If $f$ were continuous, I would try the solution here, but I can't because there is no statement that $f$ is continuous. What should I do?
 A: So your first condition is widely known as lower semi-continuity. A variant of extreme value theorem applies to such functions, see this: 
Any lower semicontinuous function $f: X \to \mathbb{R}$ on a compact set $K \subseteq X$ attains a min on $K$. 
However you need a compact domain. To achieve that in your case you have to realize that the condition 
$$\lim_{\lVert x\rVert->\infty}f(x)=1$$
implies that there's a well defined (and still lower semi continous) induced function
$$F:\mathbb{S}^n\to\mathbb{R}$$
$$F(x)=f(x)$$
$$F(\infty)=1$$
where $\mathbb{S}^n$ is an $n$-dimensional sphere constructed as the one-point compactification of $\mathbb{R}^n$, i.e. $$\mathbb{S}^n=\mathbb{R}^n\cup\{\infty\}$$
This space is compact and thus $F$ has to achieve its minimum somewhere, possibly in $\infty$.
Now since $$F(\infty)=1\geq f(c)=F(c)$$
for some $c$ then we see that if minimum is in $\infty$ then $F(\infty)=F(c)$ and thus $c$ is a minimum as well. All in all the minimum has to be in $\mathbb{R}^n$ and thus $f$ has minimum as well (since it shares values with $F$). $\Box$
