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anonymous
 one year ago
here sir
anonymous
 one year ago
here sir

This Question is Closed

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I dont understand the last sentence.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I know what Hausdorff is.

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 U_{x} and U_{y} of x and y respectively that are disjoint. This implies X is Hausdorff with these properties except one.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2It gives a definition of Hausdorff, and then said it implies something. There is no question here.

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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2that says the following: For all x and y in some space x, x is related to y.

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

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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Are you listening to me?

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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2ok man, I am gonna go then... this is pointless and a waste of time.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they said i should bring out the odd one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is an option question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am only trying to list all the options sir

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2for the 5th time, what is the question?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2and please dont just retype what you already typed...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they defined the Hausdorff space and gave some properties , asking me to point out the odd property

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2do you mean \(U_x \bigcap U\{y\} = \phi\) ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2or \(U_x \bigcap U_y\{y\} = \phi\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes and i know that option A and B are properties of Hausdorff space but the C and D is where u am confused

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, i think option C is a typo error

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Is there a reason you are not using parentheses when I keep asking you to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0may be they wanted to state\[ U_x \bigcap U_y = \phi \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(\forall x, y \epsilon\mathbb X; x \neg y\) makes no sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sir, i am so sorry about the parentheses but that was how the question came

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2But then you say yes when I say "should it be like this"

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2ok they all make sense except the equivalence one. \(\forall x, y \epsilon\mathbb X; x \neg y\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I again, don't see how you are learning here. but what ever...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what about the last option. is it a property of Hausdorff space?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean this option \[∀x,yϵX,x⋂y=0 \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2sort of, it should say \(x\ne y\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(∀x,yϵX\text{ where } x\ne y, \{x\}⋂\{y\}=0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so sorry again sir for the parenthesis , its because that the way it came . i am so sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A topological space X satisfies the first separation axiom T_{1 } if each one of any two points of X has a nbd that does not contain the other point. Thus,X is called a T_{1}  space otherwise known as what?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think it is Hausdorff Space. am i correct ? @zzr0ck3r

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2no, it is not necessarily Hausdorff. I think they sometimes call \(T_1\) "accessible space". But I have never heard it referred to as anything but \(T_1\).
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