Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

You're asking how to classify a line segment that doesn't contain its endpoints if it's embedded in R2?
kinda yes!!
Are you okay with why it's definitely not open?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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 about it?
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question