# Compact Sets, open covers

• Dec 17th 2009, 10:07 AM
p00ndawg
Compact Sets, open covers
Show that each subset of R is not compact by describing an open cover for that has no finite subcover.

A) [1,3)

C) { 1/n : n in N}

What I dont understand is when my professor did a few examples of these, he would often, in for example [0, 1), write an infinite union of sets but would start it slightly less than 0?

For example for A would, $\cup^\infty$ (0, 3/n) work?
• Dec 17th 2009, 10:29 AM
Plato
Quote:

Originally Posted by p00ndawg
Show that each subset of R is not compact by describing an open cover for that has no finite subcover.
A) [1,3) C) { 1/n : n in N}
For example for A would, $\cup^\infty$ (0, 3/n) work?

No that does not work for part A. But $\cup^\infty$(0, 3-(1/n)) will work.

For part C, notice that $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$.
• Dec 17th 2009, 10:36 AM
p00ndawg
Quote:

Originally Posted by Plato
Yes that works for part A.

For part C, notice that $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$.

why is it that you want to go slightly over or slightly below the set in question?

is it to ensure that the open cover, actually covers the set?
• Dec 17th 2009, 10:38 AM
qspeechc
I'll deal with this first

Quote:

Originally Posted by p00ndawg
What I dont understand is when my professor did a few examples of these, he would often, in for example [0, 1), write an infinite union of sets but would start it slightly less than 0?

You need to use open sets, e.g. intervals of the form (a,b), or unions thereof. So if each set you are using in the open covering is an open set, how can you cover the left-hand end of the interval [0,1) by open sets? can you use (0,1)? Or (0.1,1)? You have to have at least one set that's something like (-0.1,1), do you see why? Because the point 1 must be contained in one of the open sets in the covering, but 1 is not in the set (0,1), or (0.1,1), but is in the set (-0.1,1).

Quote:

For example for A would, (0, 3/n) work?
This would not work for A), if this is what you're asking, for the same reason as given above. This is not a covering of [1,3) because the point 1 is not contained in any of the sets (0,3/n). But you almost have the correct idea here, modify it slightly in light of what has been said before. Then check that 1) you have an open covering that 2) is the union of infinitely many sets, and 3) has no finite sub-covering, i.e. you cannot choose finitely many of the sets making the covering and get another open covering of the interval [1,3) of this finite number of open sets (Hint: to show this try proof by contradiction; what if there were a finite sub-covering?).
EDIT: there's an error, (1,3/n) would not work, I mis-read what you wrote. Ask yourself, would (0,3/n) work?
• Dec 17th 2009, 04:10 PM
platinumpimp68plus1
The idea is that, when an interval is half open, it doesn't contain its limit point. This means that there are infinitely many points approaching the point (in this case, 0 and 3), but never actually reaching it. No matter how arbitrarily close you get to the limit point, you'll always be missing some points and require additional subcovers. (ie. an infinite amount)

For 1/n, (which is the interval (0,1]) we have {1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7,......} etc., the numbers will keep getting closer and closer to 0 but never reach it. An infinite open cover of {1/n} is (1/n, 1).
• Dec 17th 2009, 09:51 PM
Drexel28
Quote:

Originally Posted by platinumpimp68plus1

For 1/n, (which is the interval (0,1]) .