Real Analysis Intervals and Boundedness Let $I:=[a,b]$ and let $f:I\to\Bbb R$ be a (not necessarily continuous) function with the property that for every $x \in I$, the function $f$ is bounded on a neighborhood $V_{\delta_x}$ of $x$. Prove that $f$ is bounded on $I.$
I am not entirely sure where to start here. Can I say that the $\lim x_n = x$ and so there exists $N_1$ in the natural numbers such that $x_n$ is in the neighborhood for any $n>N_1$ and then continue?
Please help!
 A: Suppose that $f$ is not bounded on $I$. Then for each $n\in\Bbb N$ there is an $x_n\in I$ such that $|f(x_n)|\ge n$. The sequence $\langle x_n:n\in\Bbb N\rangle$ is bounded, since it lies in $I$, so it has a convergent subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$. Let $x$ be the limit of this subsequence. There is an $m\in\Bbb N$ such that $x_{n_k}\in V_{\delta_x}$ for every $k\ge m$, contradicting the boundedness of $f$ on $V_{\delta_x}$.
Added: Recall that the points $x_n$ were chosen so that $|f(x_n)|\ge n$ for each $n\in\Bbb N$. In particular, $|f(x_{n_k})|\ge n_k$ for all $k\in\Bbb N$. Now let $M$ be any positive real number. There is a $k_0\in\Bbb N$ such that $n_{k_0}\ge M$. (Recall that $\langle x_{n_k}:k\in\Bbb N\rangle$ is a subsequence of $\langle x_n:n\in\Bbb N\rangle$, so $\langle n_k:k\in\Bbb N\rangle$ must be strictly increasing.) Now choose any $k\ge\max\{m,k_0\}$; then $n_k\ge n_{k_0}\ge M$, and $x_{n_k}\in V_{\delta_x}$ (because $k\ge m$), so $x_{n_k}$ is a point of $V_{\delta_x}$, and $|f(x_{n_k})|\ge M$. $M$ was arbitrary, so there are points of $V_{\delta_x}$ at which $f$ assumes values with arbitrarily large magnitudes $-$ which is exactly what it means for $f$ to be unbounded on $V_{\delta_x}$.
A: If you are familiar with compactness, for every $x$ let $V_x$ be an appropriate neighbourhood of $x$, and $B_x$ an associated bound.  Then the $V_x$ form an open cover of our interval $[a,b]$. There is therefore a finite subcover $\{V_{x_1},\dots, V_{x_n}\}$. Let $B=\max B_{x_i}$. 
