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?

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Hint: $I$ is a compact set. What can you say about the collection $\{V_{\delta_x}\}_{x\in I}$ in relation to $I$? – J. Loreaux Oct 23 '12 at 18:45
We have not yet used compact sets in class. – Jackson Hart Oct 23 '12 at 18:49
Have you yet covered the fact that every bounded sequence of reals has a convergent subsequence? – Brian M. Scott Oct 23 '12 at 18:53
Yes, the B-W theorem. Could I say that Suppose that f is unbounded. Then there exists xn∈[a,b] with (for example) limn→∞f(xn)=∞. Because the sequence xn is bounded, it has a convergent subsequence, denote it also xn→x0∈[a,b]. Then use the fact that f is bounded near x0 to derive a contradiction. – Jackson Hart Oct 23 '12 at 18:56
Precisely so, Jackson. – Cameron Buie Oct 23 '12 at 18:59

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 $V_{\delta_x}$.

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It should be noted that $x\in I$ since $I$ is closed. – J. Loreaux Oct 23 '12 at 19:00
I understand everything here except the part that says |f(xn)|≥n. – Jackson Hart Oct 23 '12 at 19:04
Is it like that simply because it isn't bounded? – Jackson Hart Oct 23 '12 at 19:06
@Jackson: Yes: if it’s not bounded, it assumes values with arbitrarily large magnitudes. In particular, for any $n\in\Bbb N$ there must be some point at which $f(n)\ge n$ or $f(n)\le -n$, i.e., $|f(n)|\ge n$. – Brian M. Scott Oct 23 '12 at 19:09
Ok, I get that now. I am still slight confused for some reason. Could you help explain to me how this contradicts the boundedness of f on V? – Jackson Hart Oct 23 '12 at 19:12

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 $B=\max B_{x_i}$.

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We have not learned yet about compactness and so I am pretty sure that our teacher will not let us use that. – Jackson Hart Oct 23 '12 at 18:49