A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
here
anonymous
 one year ago
here

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let R be endowed with the usual standard topology. Consider Y = [1,1] as a subspace of R. Which one of the following sets is closed in Y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0{x: \[\frac{1}{2}\ < \] x < 1}

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0{x:\[ \frac{1}{2} \] <\[ x\leqslant \]1}

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2yes, since it is open in R and and it equals the intersection with the subspace.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[{x: \frac{1}{2}\leqslantx < 1} \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2neither open nor closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[{x: \frac{1}{2}\leqslantx < 1} \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2thats the same. please use parentheses. bbl

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[{x: \frac{1}{2} < x\leqslant1} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[{x: \frac{1}{2}\leqslantx\leqslant1} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i presume this is close

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2closed sets in a subspace S, are of the form \(A\cap S\) where \(A\) is open in the parent topology.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2ok walking out the door

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and which is open in Y with same options ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it \[{x: \frac{1}{2}\ < x < 1} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02 With the standard topology on R,which one of the sets in question (1) above is open in R? with same options given above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0With the standard topology on R,which one of the sets in question (1) above is closed in R? with same options given above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sir, i am here @zzr0ck3r

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, can you help with the asked question sir?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\((1, \dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\) is open \((1, \dfrac{1}{2}]\cup [\dfrac{1}{2}, 1)\) is not open \([1, \dfrac{1}{2})\cup (\dfrac{1}{2}, 1]\) is not open \([1, \dfrac{1}{2}]\cup [\dfrac{1}{2}, 1]\) is not open

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so which means that \[ x:1/2 <x<1 \] is open in the usual standard toplogy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about in the standard topology on R

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02 With the standard topology on R,which one of the sets in question (1) above is open in R with same options given above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hello sir, @zzr0ck3r

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me post the question in full . please don't get mad at me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0With the standard topology on R,which one of the sets is open in R?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ {x: \frac{1}{2}\ < x < 1} \] \[{x: \frac{1}{2} < x\leqslant1} \] \[{x: \frac{1}{2}\leqslantx < 1} \] \[{x: \frac{1}{2}\leqslantx\leqslant1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the answer still the first here sir?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i guess from what you thought, A is the only open set while B,C, D are close sets

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A topological space is called a Hausdorff space, if for each x, y of distinct points of X, there exists nbds \[U_{x} \] and \[U_{y} \] of x and y respectively that are disjoint. This implies X is Hausdorff wiith these properties except one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A) if \[ \forall x, y \epsilon\mathbb X; x \neg y \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0B)\[There \exists U_x \epsilon N(x). U_y \epsilon N(y) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0C) \[U_x \bigcap Uy = \phi \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0D)\[\forall x, y \epsilon X, x \bigcap y = 0 \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what is the question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should i post it in anew question?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.