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I would like a to ask a question about the solution on Problem 2.17 from "Priniciples of Mathematical Analysis" by Walter Rudin. The solution is in the manual by Roger Cooke.

The question is as follows: Let E be the set of all $y \ \epsilon \ [0,1]$ whose decimal expansion contains only the digits 4 and 7. Is E compact?

For this purpose the solution shows that E is closed (as it is obviously bounded). It shows that the complement of E ($E^c = [0,1] \setminus E$) is open by proving that every point of $E^c$ is an interior point. The condition to prove this is in the end $$ |x-y| \ge \frac{1}{9*10^{n}}$$ where $x \ \epsilon \ E$ , $y \ \epsilon \ E^c$ and $n$ the first digit in $x$ which contains a number different from 4 or 7.

From this, the proof concludes that every point of $E^c$ is an interior point. It is this very last step which I do not understand. To show that x is an interior point - in my thinking - we have to show that there exists a neighbourhood arround x which contains only points that are $ \epsilon \ E^c$ as well. But the above displayed condition does not guarantee this. Since it depends on $n$, I can always find a $y \ \epsilon \ E$ which is closer to x by choosing a y which differs from x in the n+1-th digit for the first time (instead of th n-th digit). So how can x be an interior point?

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  • $\begingroup$ I think that there is some confusion between $x$ and $y$ (for example, $x\in E$ doesn't have any digit which is different from $4$ or $7$). $\endgroup$ – Riccardo Ceccon Sep 2 '18 at 8:13
  • $\begingroup$ For $x\in [0,1]$ let $f(x,n)$ be any possible value of the $n$th decimal place of $x$ to the right of the decimal point. Suppose $S\subset [0,1]$ and $\forall x\in S\;\forall n\; (x_n\not \in \{0,9\} .$ If $(x_m)_m$ is a sequence of members of $S,$ converging to $x,$ then $(f(x_m,n))_m$ converges to $f(x,n) $ for each $n.$ So $(f(x_m,n))_m$ is an eventually-constant sequence for each $n.$ Therefore if $S\subset [0,1]$ and $f(y,n)\in \{4,7\}$ for all $y\in S$ then any limit point $x$ of $S$ will satisfy $f(x,n)\in \{4,7\}$ for all $n.$ $\endgroup$ – DanielWainfleet Sep 2 '18 at 9:47
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It should read that $n$ is the first digit in $y$ that is different from $4$ or $7$ (after all the $y$ is the one in the complement of $E$). Thus $n$ doesn't vary with $x$ and for a given $y\in E^c$, all $x\in E$ are greater than a certain distance away.

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So in short he shows that if $x \in E$, $y \notin E$, then there is some $r_y >0$ (that depends on $y$, as defined by the formula, defined from the first expansion digit of $y$ that's not $4$ or $7$, which exists as $y \notin E$) such that $d(x,y) \ge r_y$.

Now if we fix $y \notin E$ for now, the ball $B(y, r_y)$ does not contain points of $E$, because points in the ball have distance $<r_y$ to $y$ while points of $E$ have distance at least $r_y$ as we saw. But this says that $B(y, r_y) \subseteq E^c$. As this holds for all $y \in E^c$, all points of $E^c$ are interior points and so $E^c$ is open and so $E$ is closed.

For me, it would be easier to show that all sets $E(n,i)$, ($n \in \mathbb{N}, i \in \{0,1,\ldots,9\}$) of those points of the reals that have value $i$ at the $n$-th place of their digit-expansion are closed, and note that your $E$ equals $\bigcap_{n \in \mathbb{N}} (E(n,4) \cup E(n,7))$ which is an intersection of closed sets and hence closed.

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