Topology of the Real Numbers Question?

Question

I've been really struggling with this question.
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Let {[x_j,y_j]}_(j>=0) be a sequence of closed, bounded intervals in R, with x_j<=y_j for all j>=1.

Suppose that the intervals which make up this sequence are disjoint, i.e. [x_j,y_j]∩[x_k,y_k] = Ø for j≠k.

Is it possible that R=U_(k>=1) [x_k,y_k]?

Hint: What sort of subset in R must be then be formed by the endpoints U_(k>=1) {x_k,y_k}
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Thank you so much!

Answer 1

Okay, I *think* I have it. Of course, I've thought that a few times now. :-) Let's suppose that the proposition is true, and that R = U{[x_j, y_j]}.

First, I can say that part of the answer to your hint is that the set of endpoints--let's call it E--is closed; this is obvious, since R\E is the union of the open intervals (x_j, y_j), and hence is itself open.

But E is more than that. To see this, pick any x_j in E, and consider the set Yj = {y_k: k in N} ∩ (-∞,x_j). (In other words, Yj is all of the y_k's that are less than x_j.) Note that Yj is a subset of E, and is bounded above by x_j; since E is closed, by the least upper bound property Yj has a least upper bound L in E. But clearly L = x_j, for if not, then the interval (L,x_j) is uncovered by the set of closed intervals.

Since x_j is not in Yj, this means that x_j is a limit point of Yj, and hence of E. By a similar argument, every y_j is also a limit point of E. Therefore, E isn't just closed--it's perfect! But, perfect sets in R are uncountable [see, e.g. Rudin's Principles of Mathematical Analysis, theorem 2.43--at least in the 3rd edition], while E is countable!

Therefore we have a contradiction, and so R *cannot* be covered by such intervals.
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It occurred to me after the fact that there is a much simpler way to show that every point in E is a limit point.

Let x_j in E, and let m in N. Then there must be a y_j_m in (x_j - 1/m, x_j), else that interval is uncovered by the set of closed intervals. Thus the sequence {y_j_m: m in N} converges to x_j, and so x_j is a limit point of E. A similar argument holds for any y_j.
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Answer 2

complex step. lookup on to the search engines. this will help!