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Let’s assume we have a function called $f$ with the domain $(0, 1)$, and let’s also assume that we replace any number like $a + 0.99999...$ with $a + 1$.

If $f$ is defined like this:

$$f(0.a_{1}a_{2}a_{3}...) = 0.a_{1}0a_{2}0a_{3}0...$$

What are the discontinuous and continuous points of $f$? How can we prove the answer?

Jigsaw
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3 Answers3

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Short, intuitive answer:

If $x$ has infinite decimal expansion, you can change it as far back in that expansion as you want, and the value of $f(x)$ will only change (twice as) far back in its expansion.

If $x$ has terminating decimal expansion, then even the smallest decrease in $x$ must change the expansion of $x$ at the terminating digit (or earlier). This turns your to give a jump in the value of $f$.

Arthur
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  • Could you elaborate the terminating digits case? – Milten Oct 30 '19 at 07:22
  • So you mean that the $f$ provided in the question is only continuous at irrational numbers? (Because non terminating decimals are irrationals) – Jigsaw Oct 30 '19 at 07:37
  • @Milten Sure. Here is an example: with $x=0.5$, we get $f(x)=0.5$. However, for $x=0.4999$, say, we get $f(x)=0.4090909$. The fact that you get $409$ instead of $49$ gives you a discontinuity. – Arthur Oct 30 '19 at 07:39
  • @Jigsaw No. It is also continuous at $\frac13$. – Arthur Oct 30 '19 at 07:42
  • @Arthur I see, thanks. – Milten Oct 30 '19 at 07:47
  • @Arthur So what is the mathematical difference here between 0.5 and 0.4999 that makes one continuous and the other one discontinuous based on your comment? (Both are terminating decimals) – Jigsaw Oct 30 '19 at 07:54
  • @Jigsaw I never said is was continuous at $0.4999$. I used $0.4999$ as an example to demonstrate that while $f(0.5) = 0.5$, the moment you decrease that input by any miniscule amount, the function value drops by more than $0.09$, no matter how small change in input you make. This makes it discontinuous at $0.5$. Of course, the function is disctontinuous at $0.4999$ as well, for basically the same reason (this time the corresponding drop in function value is $0.00000009$ instead, though). – Arthur Oct 30 '19 at 08:07
  • @Arthur so what are the continuous points, after all? – Jigsaw Oct 30 '19 at 08:11
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    @Jigsaw I said so in my answer. Any number with non-terminating decimal expansion. – Arthur Oct 30 '19 at 08:15
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This is more of a comment/addendum to Arthur's answer, which is however much to long for a comment. I think the argument in case of the non-terminating decimal expansion needs to be exapnded a bit.

The thing is, if you have say $x=0.50\ldots0\overline{1234}$, and there are $1000$ zero digits between the initial $5$ and the periodical part, then decreasing the number by $\Delta x=10^{-1001}$ will give $x-\Delta x=0.49\ldots9\overline{1234}$ and $f(x-\Delta x)=0.409\ldots$, that means the very small change by $\Delta x$ caused a very big change in $f$.

The argument goes that for a non-terminating expansion of $x$, for each integer $n$ you can find an integer $N > n$ such the the $N$-th decimal digit of $x$ $d_N$ is neither $0$ nor $9$. That means if you change $x$ by $\Delta x$ with $\lvert\Delta x\rvert < 10^{-N}$, you are changing $d_N$ by at most $1$ (in either direction), and since $d_N \neq 0,9$, that cannot cause an overcarry into the previuous digit (if $x$ was increased) or an undercarry into the previous digit (if $x$ was decreased).

So in either cases, the decimal expansion of $x+\Delta x$ will be the same as for $x$ up to at least the $(N-1)$-th decimal, so their $f$ values will be the same at least until the $(2N-2)$-th decimal digit, as Arthur said.

Ingix
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Consider $$f_n(x) = \frac{\{10^{n}x\}}{10^{2n}}-\frac{\{10^{n+1}x\}}{10^{2n+1}},\ \ n=0,1,2\dots,$$ where $\{ x\}$ is fractional part function. Note that $f_n(x)$ is right-continuous everywhere and continuous at all non terminating (rational or irrational) decimals.

Now, your $f(x)$ is just

$$f(x)=\sum_{n=0}^{\infty}f_n(x).$$

The series converges unformly, by Weierstrass M-test. Since uniform convergence preserves continuity, we conclude that $f(x)$ too is right-continuous everywhere and continuous at all non terminating decimals.

Generalization of Arthur's example in comments leads to the conclusion that the function is not left-continuous at terminating decimals numbers.

You might want to have a look at this similar function, where all continuities and discontinuities are dealt with using decimal expansions.

dfnu
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