A question on an unbounded function Does there exist a function $f$ such that it has a finite value for each point $x$ of $[0,1]$, however for any nbhd of $x$ $f$ is unbounded?
Thanks for your help.
 A: I think, such function can be given by 
$$
  f(x) = \begin{cases}
0,\text{ if }x\notin\Bbb Q
\\
n,\text{ if }x = m/n \in \Bbb Q,
\end{cases}
$$
where the representations of rationals as fractions is assumed to be irreducible.
Clearly, the function is finite (point-wise). To show that it is unbounded consider the following argument.
If $x$ is irrational, you can consider the sequence $x_k\to x$ where $x_k$ is a decimal representation of $x$ up to the $k$'th digit. It shall be clear, that $f(x_k)$ grows unboundedly: there will be an infinite number non-zero digits in the decimal representation of $x$, so that $f(x_k)$ contains an infinite number of powers $10^k$.
If $x$ is rational, when you fix a neighborhood of it you're interested in, pick there any irrational number $x'$ and repeat the procedure above.
A: You could try the Conway Base $13$ Function, which takes every value in $\mathbb R$ in the neighbourhood of any point. It has the intermediate value property, but is nowhere continuous.
The basic idea is to use the base $13$ expansion of a number as code for a decimal (base $10$ expansion), with the extra symbols coding for $+$, $-$ and the decimal point. "Impossible" numbers map to zero, but if the trailing part of the base $13$ expansion is a code for a decimal number, then you map to that.
