# Very basic topology question

• Jun 15th 2011, 09:18 PM
BWilson
Very basic topology question
Hello, I am having some difficulty on problem 17 of chapter 2 in Rudin's Principles of Mathematical Analysis. The problem reads: "Let E be the set of all x in [0,1] whose decimal expansion contains only the digits 4 and 7. Is E compact?"

My problem is that there is a theorem in this book which says that if every infinite subset of a set E has a limit point in E, then E is compact.

I used the following to justify that E is compact:
By the way, I denote the neighborhood of a point, p, of radius r as N(p, r).

Let F be an infinite subset of E and let s be a finite set of 4's and 7's with m digits. Since F is infinite, F contains every combination of n 4's and 7's, so s is an element of F.
Then there exist an s1 and s2 such that |s1 - s2| < 3 X 10^-m.
s1 and s2 are therefore limit points since s1 is in N(s2, x) and s2 is in N(s1, x) for any x > 3 X 10^-m. This x can always be found since m is a natural number with no upper bound. Therefore, every infinite subset of E contains a limit point in E, and E is compact.

However, the following seems to be an open cover of E for which there is no finite subcover:
F(n) = (0, .7) where F(1) = (0, .7), F(2) = (0, .77) F(3) = (0, .777), and so on. This would seem to imply that E is not compact .

I'm sure I'm making some silly mistake, but I would really appreciate it if someone could help me sort out what I'm doing wrong.
• Jun 15th 2011, 11:16 PM
Drexel28
Re: Very basic topology question
Quote:

Originally Posted by BWilson
Hello, I am having some difficulty on problem 17 of chapter 2 in Rudin's Principles of Mathematical Analysis. The problem reads: "Let E be the set of all x in [0,1] whose decimal expansion contains only the digits 4 and 7. Is E compact?"

My problem is that there is a theorem in this book which says that if every infinite subset of a set E has a limit point in E, then E is compact.

I used the following to justify that E is compact:
By the way, I denote the neighborhood of a point, p, of radius r as N(p, r).

Let F be an infinite subset of E and let s be a finite set of 4's and 7's with m digits. Since F is infinite, F contains every combination of n 4's and 7's, so s is an element of F.
Then there exist an s1 and s2 such that |s1 - s2| < 3 X 10^-m.
s1 and s2 are therefore limit points since s1 is in N(s2, x) and s2 is in N(s1, x) for any x > 3 X 10^-m. This x can always be found since m is a natural number with no upper bound. Therefore, every infinite subset of E contains a limit point in E, and E is compact.

However, the following seems to be an open cover of E for which there is no finite subcover:
F(n) = (0, .7) where F(1) = (0, .7), F(2) = (0, .77) F(3) = (0, .777), and so on. This would seem to imply that E is not compact .

I'm sure I'm making some silly mistake, but I would really appreciate it if someone could help me sort out what I'm doing wrong.

The thing though is that your open cover doesn't contain $\displaystyle .\overline{7}$ which it approaches but which is also in the set. Secondly, your proof looks fine to me but I would have appealed to the Heine-Borel theorem for a quick finish (if you mean compact as a subspace of $\displaystyle [0,1$ there is still no issues since a space is compact in a subspace implies that it is compact in the ambient space).
• Jun 16th 2011, 01:16 AM
ojones
Re: Very basic topology question
The best way to show that $\displaystyle E$ is compact is to show that it is closed and bounded. It's obviously bounded. For closure, start by letting $\displaystyle p$ be a limit point of $\displaystyle E$ and showing that $\displaystyle p\not \in E$ leads to a contradiction.
• Jun 16th 2011, 08:12 AM
Deveno
Re: Very basic topology question
Quote:

Originally Posted by BWilson
Hello, I am having some difficulty on problem 17 of chapter 2 in Rudin's Principles of Mathematical Analysis. The problem reads: "Let E be the set of all x in [0,1] whose decimal expansion contains only the digits 4 and 7. Is E compact?"

My problem is that there is a theorem in this book which says that if every infinite subset of a set E has a limit point in E, then E is compact.

I used the following to justify that E is compact:
By the way, I denote the neighborhood of a point, p, of radius r as N(p, r).

Let F be an infinite subset of E and let s be a finite set of 4's and 7's with m digits. Since F is infinite, F contains every combination of n 4's and 7's, so s is an element of F.
Then there exist an s1 and s2 such that |s1 - s2| < 3 X 10^-m.

i don't think this is true. let F = {0.7, 0.77, 0.777, 0.7777,....etc.}. then F is an infinite subset of E, but if s is a decimal that contains ANY 4's it is not in F.

Quote:

s1 and s2 are therefore limit points since s1 is in N(s2, x) and s2 is in N(s1, x) for any x > 3 X 10^-m. This x can always be found since m is a natural number with no upper bound. Therefore, every infinite subset of E contains a limit point in E, and E is compact.
what set are s1, s2 supposed to belong to? E? F? s1, s2 are limit points of what set? i find this hard to follow. presumably you mean F. the fact that F is infinite does not mean F contains every finite combinations of 4's and 7's. it may not contain any elements with 4's in their expansion, or it may contain just a single one.

i understand the "spirit" of what you mean to say. you want to show that any infinite subset F of E has a limit point in E. what i think you want to do is for every m in N,

find some x_m in E that is within 4 x 10^(-m) of a point in F. then the sequence {x_1, x_2,....,x_m,...} converges to a point in F, so F has a limit point in E,

namely lim m-->∞ {x_m}. why do i use 4 x 10^(-m) instead of 3 x 10^(-m)? because 7/9 - 4/10 = 34/90 = 0.377.... so at stage 1, our first choice for x_1

might be 7/9, whereas the limit point we are searching for in F might be 0.4444.... , so a neighborhood of radius 0.3 isn't "big enough", and we want to proceed

"digit-by-digit". we have no idea where the points of F might be "clustered", and we want to be sure our successively smaller neighborhoods of our (somewhat arbitrary)

elements of E actually contain x_j at the j-th step.

Quote:

However, the following seems to be an open cover of E for which there is no finite subcover:
F(n) = (0, .7) where F(1) = (0, .7), F(2) = (0, .77) F(3) = (0, .777), and so on. This would seem to imply that E is not compact .

I'm sure I'm making some silly mistake, but I would really appreciate it if someone could help me sort out what I'm doing wrong.
as drexel28 pointed out above, this is not a cover.

now, the question remains, how do we know that the succesive neighborhoods N(x_1, 0.4), N(x_2, 0.04), etc. all contain points of F?

well that is where the infiniteness of F comes in. the first neighborhood covers all of E (so it has to cover F), no matter what our choice of x_1 is.

for our second neighborhood, x_2 has to either be in [44/100 43/90] or [74/100 7/9] (we can safely exclude 0.4 and 0.7 as limit points of F).

one of these sets has to be infinite (perhaps both are) so we pick an infinite one (call it F_2) and select x_2 (arbitrarily) from E∩F_2.

(you may notice we are mimicking the argument for bolzano-weierstrass). *waving hands* continuing in this way.....

(we will have 4 subintervals to consider on the 3rd round:

0.444 to 0.4477777...
0.474 to 0.4777777...
0.744 to 0.7477777...
0.774 to 0.7777777... one of these will have to contain an infinite number of elements of F, and the size of each interval is < 0.004).
• Jun 16th 2011, 08:21 AM
Tinyboss
Re: Very basic topology question
Another way, using the limit point compactness property you mention (LPC implies compact in metric spaces):

Let S be an infinite subset of E. We will produce a limit point s of S. Begin at the first digit after the decimal--since S is infinite, there are either infinitely many elements of S with first digit 4, or infinitely many with first digit 7...either way, record either 4 or 7 as the first digit of s, and proceed by induction. If we have produced N digits of s, then there are infinitely many elements of S with the same first N digits, and either infinitely many have a 4 next, or infinitely many have a 7 next. So we may produce infinitely many digits of s, which specifies s, and clearly s is in S since it has only 4s and 7s in its expansion.

Notice how this parallels the standard proof of Heine-Borel, which might be Rudin's aim in writing the problem.
• Jun 16th 2011, 09:03 AM
BWilson
Re: Very basic topology question
Quote:

Originally Posted by Drexel28
The thing though is that your open cover doesn't contain $\displaystyle .\overline{7}$ which it approaches but which is also in the set. Secondly, your proof looks fine to me but I would have appealed to the Heine-Borel theorem for a quick finish (if you mean compact as a subspace of $\displaystyle [0,1$ there is still no issues since a space is compact in a subspace implies that it is compact in the ambient space).

Thanks for the helpful response. I must have been thinking about an infinite set which was closed on that end for some reason. Thanks also for the Heine-Borel theorem explanation.
• Jun 16th 2011, 09:36 AM
BWilson
Re: Very basic topology question
Quote:

Originally Posted by Deveno
i don't think this is true. let F = {0.7, 0.77, 0.777, 0.7777,....etc.}. then F is an infinite subset of E, but if s is a decimal that contains ANY 4's it is not in F.

You're right there. Also, s1 and s2 were meant to be in F. I think I was trying to make the proof more general than it has to be. You're method of proving seems good, but going off the one I had already made, would it be acceptable to change the neighborhood to 100^ -m and to state that for some combination of 4's and 7's called s, s is in F? (I had previously tried to say that every such s was in F). This would provide a neighborhood between any sequence of m 7's or 4's and the same sequence of 7's and 4's with the last digit removed. Naming these two points s1 and s2, s2 is in N(s1, 100^-m) and s1 is in N(s2, 100^-m).