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Let $E$ be the set of all $x \in [0,1]$ whose decimal expansion contains only the digits 4 and 7. Is $E$ compact?

This question comes from Baby Rudin Ch.2 - 2.17.

I have a solution I found for the question and this is what my questions revolve around. I believe I have the correct understanding of what was set out to be accomplished in order to prove the claim (I will type my "understanding" below). But I am a little lost in how the bounds that were stated were derived.

As such here is the solution:

enter image description here

My understanding: In order to show that $E$ is compact we have to show closed and boundedness of the set. Boundedness is obvious, to show it is closed we show that $E^{c} = [0,1] -E $ is open. This is accomplished by showing that any point $y \in E$ will be at the minimum a fixed distance away from every point of $E^{c}$ which we can then use to choose $\epsilon$ to always be less than this distance. which will show that $E^{c}$ is open (this is a new technique to show openess that I've seen). But after concluding $E^{c}$ is open then we can conclude $E$ is closed.

Questions:

i) How are these bounds constructed? i.e how is $ \epsilon \leq \frac{7}{9} 10^{-m}$ found? And as such how do the other two bounds follow? i.e: $\frac{2}{9 \cdot 10^{m}}$

ii) As mentioned above is this another wy to show a point is an interior point? i.e: by showing no point from outside the set could be a fixed distance away?

iii) How is it we are allowed to pick our $\epsilon$ in these sorts of questions? Is it because we are trying to establish a fact about a certain collection of elements and as such this $\epsilon$ works only for this set?

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    $\begingroup$ Just observing that $[0,1]\cong\{0,1,\ldots,9\}^{\omega}$ and your subset is $\{4,7\}^{\omega}$. $\endgroup$ – Inactive - avoiding CoC Mar 10 at 2:34
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    $\begingroup$ I would prove the claim by recognizing your set as the intersection of (infinitely many) closed sets. $\endgroup$ – Lubin Mar 10 at 2:40
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This is a simple application of a geometric sum; as noted $|x_k-y_k|\leq7$ and so $$\varepsilon=\sum_{k\geq m+1}10^{-k}|x_k-y_k|\leq7\sum_{k\geq m+1}10^{-k}=\frac{7}{9}10^{-m}.$$ Then plugging this back into the earlier inequality yields $$|x-y|\geq10^{-m}-\varepsilon=\frac{2}{9}10^{-m}.$$

As for the structure of the proof; the proof shows that $E$ is closed by showing that $E^c$ is open, which by definition means that every for $x\in E^c$ there is an open interval around $x$ that is entirely contained in $E^c$.

That such an interval exists is proved by showing that for every $x\in E^c$, every element of $E$ is more than some given distance $\tfrac{1}{9\cdot10^n}$ apart from $x$, where $n$ depends only on $x$. This means the open interval $(x-\tfrac{1}{9\cdot10^n},x+\tfrac{1}{9\cdot10^n})$ is entirely contained in $E^c$.

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  • $\begingroup$ I suspected that, but when I worked it out I got: $\frac{1}{1 -\frac{1}{10}} = \frac{10}{9}$ so how does $10^{-m}$ appear ? $\endgroup$ – dc3rd Mar 10 at 2:54
  • $\begingroup$ Where did you get that numerator? The sum starts at $k=m+1$ so the numerator should be $10^{-m-1}$. $\endgroup$ – Inactive - avoiding CoC Mar 10 at 2:58
  • $\begingroup$ My issue with the expression is how does $7\sum_{k\geq m+1}10^{-k} =\frac{7}{9}10^{-m}$ when $7\sum_{k\geq m+1}10^{-k} = 7(\frac{1}{1 -\frac{1}{10}}) = 7(\frac{10}{9})$ since it is an infinite geometric series. $\endgroup$ – dc3rd Mar 10 at 3:11
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    $\begingroup$ No; are you familiar with the closed form of a geometric series? In this case $$(1-10^{-1})\sum_{k\geq m+1}10^{-k}=10^{-m-1},$$ and so the sum evaluates to $\tfrac{10^{-m-1}}{1-10^{-1}}=\tfrac{1}{9}10^{-m}$. See alse the wikipedia page; here $r=\tfrac{1}{10}$ and $a=7\cdot10^{-m-1}$. $\endgroup$ – Inactive - avoiding CoC Mar 10 at 3:14
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    $\begingroup$ $$\frac{10^{-m-1}}{1-10^{-1}}=\frac{10}{10}\frac{10^{-m-1}}{1-10^{-1}}=\frac{10^{-m}}{10-1}=\frac{10^{-m}}{9}.$$ $\endgroup$ – Inactive - avoiding CoC Mar 10 at 3:24

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