Continuity of removing odd decimals Let $x = 0.x_1x_2x_3\cdots$ be the unique decimal representation of $x$ not containing a string $999\cdots$. Let $f(x)\in (0,1)$ be $0.x_2x_4\cdots$.
Where is $f$ discontinuous?
I haven't been able to find any point and it seems to me that eventually the $n^{th}$ decimal place of a sequence has to agree with the thing it converges to so this would give a proof of continuity. I found this in a book of contest problems and it seems to easy for this to be the case.
 A: There are some discontinuities in this function. You make a slightly incorrect assertion in your reasoning that it might be continuous:

It seems to me that eventually the $n^{th}$ decimal place of a sequence has to agree with the thing it converges to so this would give a proof of continuity.

This is almost true in the sense that if a sequence $x_1,x_2,x_3,\ldots$ converges to some $x$ then the decimal expansions of $x_i$ get closer to some decimal expansion of $x$ - but possibly one that ends with $999$. For instance, consider the sequence
$$1.1, 0.99, 1.001, 0.9999, 1.00001,\ldots$$
in this example, the $n^{th}$ term has $n$ digits agreeing either with $1.000\ldots$ or $0.999\ldots$. This is an issue to be careful about whenever you see a function inspecting the digits of a number.
Your reasoning can correctly show that the function is continuous at every point not of the form $\frac{a}{10^n}$ for integer $a,n$ - that is, those points that have a unique expansion.
Points that have a terminating decimal expansion, however, are discontinuities. For instance, look near $1$. Along the sequence $0.99, 0.9999, 0.999999$, the function's values converge to $1$, but along the sequence $1.01, 1.0001, 1.000001,\ldots$ the function's values converge to $0$.
The function is not continuous at $0.1$ for similar reasons.
However, this question is tricky: $0.01$ is actually fine, however: you can use your reasoning for it with modification. Within a small enough radius of $0.001$ every decimal either agrees with $n$ digits of $0.0100000\ldots$ or $0.009999\ldots$. Agreeing with enough digits of those forces the output of the function to be close to $f(0.01) = 0.1$. The difference in behavior is because the function can actually see that $09999\ldots$ rolling over to $10000$ in a suitable way since it records the first digits of this sequence.
If you consider carefully where this argument works and where it produces counterexamples, you find that this function is discontinuous at every integer and at the set of rationals of the form $\frac{a}{10^n}$ where $a$ is an integer not divisible by $10$ and $n$ is an odd integer.
