Why a function is bounded at a neighbourhood of point c, if it's continuous at point c?

Question

I just try to understand the first Weierstrass theorem, and in the proof there it is by the way noted that a function $f$ is bounded at the neigbourhood of point $c$, if it's continuous at $c$. Can someone explain why it is so?

Answer 1

If $f $ is continuous at $c $, then by definition (of continuity at a point) $$\forall\epsilon>0 \ \ \exists \delta>0 \ \ |x-c|<\delta\implies |f (x)-f (c)|<\epsilon $$ Take for example $\epsilon =2$ (any positive $\epsilon$ will do). Then (by the above statement of continuity) for some $\delta>0$, we have $|f(x)-f(c)|<\epsilon$, or equivalently $-2<f (x)-f(c) <2$. From this we can find $K>0$ that is a bound for $f $ on the given interval.

We have $-2+f(c)<f(x)<2+f(c)$, and $K$ may be chosen as the maximum of the absolute value of these bounds (check).