Continuity on an open interval and the epsilon-delta definition of continuity

Question

The question asks:

Assume $f$ is continuous at $a$. Assume $f(a) < 4$. Prove there exists an open interval $I$, centred at $a$, such that $\forall x \in I, f(x) < 4$.

I am stuck on this question. I am supposed to use the epsilon-delta definition of "f continuous at a", which I came up as $\forall \epsilon > 0, \exists \delta > 0$ such that $|x-a| < \delta \implies |{f(x) - f(a)}| < \epsilon$. Could anyone provide me a solution or an insight as to what this question actually means?

Answer 1

I was doing this problem and found this solution:

with $4-f(a)>0$ there is $\epsilon>0$,$4-f(a)>\epsilon$.Choose $\delta>0$ with $|x-a|<\delta \implies$ $|f(x)-f(a)|<\epsilon<4-f(a) \implies -4+f(a)<f(x)-f(a)<4-f(a)$

$\implies f(x)\in(-4+2f(a),4)$

Answer 2

Suppose $f$ is as described in the problem statement and there does not exist such an interval. Let $\delta > 0$. Then there exists some $x’$ in the delta neighborhood centered at $a$ such that $f(x’) \geq 4$. In particular, let $\epsilon = {f(x’) - f(a)\over 2}$. Then for this $\epsilon$, $|x-a| < \delta$ does not imply that $|f(x) - f(a)| < \epsilon$. Thus for all $\delta>0$ there is an $\epsilon>0$ such that $|x-a| < \delta$ does not imply that $|f(x) - f(a)| < \epsilon$. But this contradicts the continuity of $f$.