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experimentX
Q1: An open interval in \( \mathbb R^1 \) is not open in \( \mathbb R^2 \), then what is it? Q2: In Cantor intersection theorem, if each Q is were non empty open set then would the intersection be empty or non-empty?
You're asking how to classify a line segment that doesn't contain its endpoints if it's embedded in R2?
kinda yes!! http://en.wikipedia.org/wiki/Open_set#Open_and_closed_are_not_mutually_exclusive
Are you okay with why it's definitely not open?
it does not contain it's end point in R^1, but it's an open set in R^1, my book says the the interval is no longer open set in R^2 because it cannot contain open two ball in R^2 space inside it.
i am guessing it to be neither open nor closed set.
I would agree. For a set to be open, there needs to exist a tiny little interval around every point in the set such that all points in the interval also belong to the set. in R1, that looks like this: |dw:1356810565391:dw|
But in R2, you need a circle, not just an interval: |dw:1356810632181:dw|
In R3 you need a sphere, etc etc etc. And the definition of a closed set is a set such that its complement is an open set, which is also false in this case because the endpoints of the interval are contained in the complement of the set, so it contains some limit points.
Circle and Sphere are actually misleading. I should have said Disk and Ball.
According to the definition on Wikipedia, this can't be clopen either.
That's right. A clopen set is both open and closed, this set is neither open nor closed.
Let's see what other people have to say ... i'll leave the post open
Any idea on Cantor intersection theorem?
what if we assume that each Qk were open and not empty? what would be the final result of the intersection of Qk's?
Then it may well be empty. Are you familiar with the proof of this theorem?
yes,, it creates a sequence from each Qk's .. and as k tends to increase to inf ,,, by bolzano Weirstrass this would be accumulation point. Since, each Qk is closed and closed set contain at least one accumulation points, at least there is one element in the intersection.
Right. If the sets are open, though, it's possible that the sequence of infima converges to an accumulation point that doesn't actually belong to the sets.