Prove that if $f(p)\in I$ then $\exists U\subset\Bbb R$ environment of $p$ such that $f(x)\in I$ $\forall $ $ x \in U$

Question

Be $f:\Bbb R \to \Bbb R$, a continuous function, $ I=(a,b)\subset\Bbb R$ an open interval. Prove that if $f(p)\in I,$ then $\exists U\subset\Bbb R$ environment of $p$ such that $f(x)\in I\forall x \in U$.

I think I can use uniform continuity, but I do not know how to do it.

Answer 1

Uniform continuity is not needed; plain and ordinary continuity at $p$ will get the job done; here I'll use the usual $\epsilon$-$\delta$ definition:

Since $I = (a, b) \subset \Bbb R$ is open and $f(p) \in I$, there is an $\epsilon > 0$ such that

$(f(p) - \epsilon, f(p) + \epsilon ) \subset (a, b) = I; \tag 1$

then since $f$ is continuous, there is some $\delta > 0$ with

$\vert f(x) - f(p) \vert < \epsilon \tag 2$

provided

$\vert x - p \vert < \delta; \tag 3$

now take

$U = (p - \delta, p + \delta); \tag 4$

it is easy to see that

$U = \{ x \in \Bbb R \mid \vert x - p \vert < \delta \}; \tag 5$

therefore, if $x \in U$, (3) binds so we have

$\vert f(x) - f(p) \vert < \epsilon; \tag 6$

then

$f(x) \in (f(p) - \epsilon, f(p) + \epsilon) \subset (a, b) = I, \tag 7$

just as we wanted to prove.