Vtyjkyuk

a) Let $f : [a, b] → mathbbR$ be a (not necessarily continuous) function with the property that, for every $x ∈ [a, b]$, there is a number $δ_x > 0$ for which $f$ is bounded on the neighborhood $Vδ_x (x)$ of $x$. Prove that the function f is bounded on the interval $[a, b]$.

b. Does the result of part a hold if the interval is $(a, b)$?

My proof for a:
Let $k = [a, b]$. If $f(k)$ is not bounded, the for all $n in mathbbN$, there exist a sequence $y_n$ is a subset of $f(k)$ such that $|y_n| > n$.
Let $x_n$ be a subset of $K$ satisfy $f(x_n) = y_n$. Then seq $x_n$ is bounded. So we extract convergent subsequences with $x_n_j$ converges to $x$ as $j$ goes to infinity. Since $K$ is closed, $x in K$. By sequence criterion for continuity, $y_n_j = f(x_n_j)$ converges to $f(x)$ . But $|y_n_j| > n_j geq j$. This is a contradiction since unbounded sequences are divergences. Thus $f(k)$ is bounded.

Is this proof good ?

For part b, could someone give some hints how to proceed this proof?

For b take the function $f : (0,1) rightarrow mathbbR$, $f(x) = frac1x$. The function is continuous in $(0,1)$. $f$ is bounded in every open nbd of $x in (0,1)$ but it is unbounded in $(0,1)$.

Proof for a is good.

For proof a you can use property of compact set in $mathbbR$.

A set $A subset mathbbR$ is said to be compact if each of its open cover has finite subcover. $[a,b]$ is a compact set in $mathbbR$.

For any $x$ in $[a,b]$ we are getting an open nbd $V_delta(x)$ of length $delta$ where the function $f$ is bounded, i.e. $exists$ $M_delta > 0$ s.t. $f(x) < M_delta forall x in V_delta(x)$. Now $V_delta(x) : x in [a,b]$ is an open cover of $[a,b]$, having a finite subcover say $V_delta(x_1), V_delta(x_2) dots V_delta(x_n)$. $[a,b] = cup_k = 1^n V_delta(x_k)$. So get $M = maxM_1, M_2, dots, M_n$. $f(x) < M$. Thus $f$ is bounded.