Bounded in $R$ equal to say bounded in every area of $a$?

Question

let $f(x)$ be a function and $a$ a real number.

I need to complete the following sentence:

______________ iff there is $\epsilon>0$ such that for every $l>0$ and every $x$, if $|x-a|<l$ then $|f(x)-f(a)|<\epsilon$

(1) $f(x)$ is bounded in $R$

(2) $f(x)$ is bounded in every area of $a$

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My question is:

Why (1) is true while (2) isn't (According to my professor in math)? don't they refer to the same thing but in different words?

Answer 1

1. Clearly, $f$ is bounded $\iff$ $f-k$ is bounded, for some (arbitrary) $k \in \mathbb{R}$. Choose $k=f(a)$. So, $f$ is bounded $\iff$ $g:=f-f(a)$ is bounded $\iff$ there exists $\epsilon >0$ such that for every $x \in \mathbb{R}$ we have $|g(x)|<\epsilon$. But this is equivalent to the presented statement, since "every $x \in \mathbb{R}$" can be obtained by "every $x$ such that $|x-a|<l$, no matter what $l>0$"

2. "$f$ is bounded" is not equivalent to "$f$ is bounded in every bounded interval around $a$: as a counterexample, choose $f(x)=x-a$. It is not bounded on all $\mathbb{R}$, but it is when restricted to, say, $[a-l,a+l]$, for an arbitrary $l>0$ (we have $|f|\leq l$)