Let $f$ be continuous on $\mathbb{R}$, and $\inf_\mathbb{R}f(x)<0$. Prove that $\exists c$ such that $f(c)<0$

Question

Problem: Let $f$ be continuous on $\mathbb{R}$, and $\inf_\mathbb{R}f(x)<0$. Prove that $\exists c$ such that $f(c)<0$.

Here's where I am: suppose $R$ is the range of $f$. Then, if $\inf_\mathbb{R} R=m$, then $\forall\epsilon > 0$ there is always one member of $R$ that is less than $m+\epsilon$. In other words, I can always find a $c$ such that $f(c) < \inf_\mathbb{R}f(x)+\epsilon$.

From here it seems intuitive that there is $c$ such that $f(x)<0$ since $\epsilon$ is arbitrarily small, so it suffices that $\epsilon<|\inf_\mathbb{R}f(x)|$, and then the equation becomes $f(c) < \inf_\mathbb{R}f(x)+\epsilon<0$. However, I can't quite help feeling this isn't a complete proof; can't $|\inf_\mathbb{R}f(x)|$ be arbitrarily small also? So then can you always find $\epsilon<|\inf_\mathbb{R}f(x)|$, in other words an arbitrarily small number that is less than another arbitrarily small number?

Also, just saying $\epsilon$ is arbitrarily small and concluding doesn't seem very rigorous to me. And none of this incorporates the fact that $f$ is continuous, which is clearly essential. Can anyone elaborate on this for me? Thanks!

Answer 1

An option:

Assume $f(x) \ge 0$ for all $x \in \mathbb{R}$.

Then $0$ is a lower bound for $f$, and

$\inf_{\mathbb{R}} (f) \ge 0$, a contradiction.

Answer 2

In your argument take $\epsilon =-\frac {\inf_{\mathbb R} f(x)} 2$. This gives $f(c)<\frac {\inf_{\mathbb R} f(x)} 2 <0$.

Answer 3

You don't need continuity.

Assume $c:= \inf_{x \in\mathbb{R}} f(x) < 0$. Choose a sequence $(x_n)_n$ with $f(x_n) \to c$. For $n$ sufficiently large, we have $f(x_n) < 0$.

Answer 4

Suppose $a = \inf{f(z) \ : z \in \mathbb{R}}$. Like a < 0 implies $\exists b$ such that $a<b<0$ and by the definition of infimum, $\exists z_{0}$ such that $a <f(z_{0})<b<0 $.