# Proving boundedness of continuous functions

If a function $f(x)$ is continuous on the closed interval $\left [ a,b \right ]$ then its bounded on this interval........the proof for this theorem i have is:

Since it's continuous on $\left [ a,b \right ]$ if we pick a random point on this interval let it be $c$

$\implies$ $\forall$ $\epsilon$ $> 0$ , $\exists$ $\delta$($\epsilon$,c) $> 0$ s.t $\left | x-c \right|$ $<$ $\delta$ $\implies$ $\left | f(x)-f(c) \right|$ $<$ $\epsilon$
$-$ $\epsilon$ $< f(x)-f(c) <$ $\epsilon$
$f(c)-$ $\epsilon$ $<$ $f(x)$ $<$ $f(c)+$ $\epsilon$

Take $M =$ $\left |f(c) \right|$ $\in$ $+$ $\mathbb{R}$
$\forall$ $M$ $\in$ $\mathbb{R}$ $f(x) < M$

$\therefore$ $f(x)$ is bounded

Is there something missing with this proof ?
because i could'nt really understand it

• @Weaam m is positive Commented Jul 26, 2017 at 19:27
• It's immediate that the proof is nonsense because you can use the same argument to "prove" the theorem for open intervals. Commented Jul 26, 2017 at 19:27

That proof in only showing that $f$ is locally bounded (bounded in a neighborhood of any given point).
The $M$ (which we better denote as $M_c$ to emphasize that it depends on the point) should really be $|f(c)|+\epsilon$ in that argument.
And the argument shows that for each $c$ there is $M_c$ and an interval $|x-c|<\delta_c$ in which the bound holds.
Taking the corresponding interval for every point, gives us a bunch of open intervals that cover all of $[a,b]$. Since this is compact, one can choose finitely many of them that still cover $[a,b]$. Now, if we take the corresponding $M$'s bounds in each of them and take the maximum, that gives a bound that does hold in all the interval $[a,b]$.
Let $f(x)=x$ on $[0,1]$ and $c=\frac12$. Then $M=|f(c)|=\frac12$. But $|f(x)| \not <M$ for all $x \in [0,1]$ (for instance, $|f(1)|>M$)