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Let $S \subset \mathbb R$ be the set of all numbers in $[0, 1]$ that have a decimal expansion that consists only of $0$’s, $1$’s, and/or $9$’s. Some examples of numbers in $S$ include $$0,\ 0.01,\ 1.0 = 0.99999999......,$$ $$1/9= 0.11111111....,\ 1/10= 0.1 = 0.09999...... ,\ 1/11= 0.09$$ as well as many irrational numbers, such as the first decimal example of a transcendental number. Decimal expansions are not unique, so $S$ also contains some numbers that you might not initially consider, such as $1/5= 0.2 = 0.1999999...$.

  1. Is $S$ open? Prove your claim.

  2. Is $S$ closed? If so, is $S$ compact? Prove your claim(s).

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  • $\begingroup$ I don't think it's open. No matter what $x\in S$ and $\epsilon>0$ you specify, one can always find a number $y$ whose decimal expansion ends $\ldots5000\dots$ far enough beyond the decimal point, s.t. $|x-y|<\epsilon$. Clearly $y\not\in S$. $\endgroup$ – Dan Apr 13 '15 at 7:54
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$T=[0,1]\backslash S$ is open. You just have to show that any element of $T$ has an open neighborhood in $T$. Notice that $T$ is the set of numbers that have no decimal expansion with only $0,1,9$ (that is, of the two possible expansions when there are two, none satisfy this constraint).

Let $x\in T$, and if there are two decimal expansions, choose the one that does not end with infinitely many $9$s. Let $x=0.x_1x_2\dots$, then there is an index $i$ such that $x_i \not\in \{0,1,9\}$.

Then two possibilities:

  1. $x_i\in\{3,4,5,6,7,8\}$, then all numbers in $]a,b[$ are in $T$, with

$$a=0.x_1x_2\dots x_{i-1}(x_{i}-1)$$ $$b=0.x_1x_2\dots x_{i-1}(x_{i}+1)$$

  1. $x_i=2$, then either there is another index $j<i$ such that $x_j\not\in\{0,1,9\}$, either there is at least one nonzero $x_j$ for $j>i$ (maybe both). Otherwise, you could replace $x_i$ with a $1$ followed by infinitely many $9$s, and $x$ would be in $S$. In both cases, you can find an index $i$ (change if necessary) such that $x_i\not\in\{0,1,9\}$ and there is a $j>i$ such that $x_j\neq 0$.
    Then you can choose $a=0.x_1x_2\dots (x_{j}-1)5$, and $b$ is the same as above.

And of course $x\in]a,b[$. Therefore $T$ is open.

Since $T$ is open, $\Bbb R\backslash T$ is closed and $S=[0,1]\cap (\Bbb R\backslash T)$ is closed. Since $\Bbb R$ is connected, $S$ can't be both open and closed (unless it's empty or equal to $\Bbb R$, which it is not), hence it's not open.

Then Heine-Borel theorem tells $S$ is compact, since it's a closed and bounded subset of $\Bbb R$.

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Direct answer on a):

Taking:

  • $r_1=0.5$
  • $r_2=0.15$
  • $r_3=0.115$
  • $r_4=0.1115$

et cetera, a convergent sequence $(r_n)$ is constructed with $r_n\notin S$ and limit $r:=0.111\cdots\in S$. This tells us that the complement of $S$ is not closed, or equivalently that $S$ is not open.

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