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anonymous
 one year ago
A topological space X satifies the first separation axiom T_{1 }if each one of any two points of X has a nbd that does not containthe other point. Thus,X is called a T_{1}  space otherwise known as ______________
Metric Space
Hausdorff Space
Frechet Space
Topological Space
anonymous
 one year ago
A topological space X satifies the first separation axiom T_{1 }if each one of any two points of X has a nbd that does not containthe other point. Thus,X is called a T_{1}  space otherwise known as ______________ Metric Space Hausdorff Space Frechet Space Topological Space

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think frechet space

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One of the conditions below is not a condition fot a topological space X to satisfy the first separation axiom If and only if all one point set in X is closed If and only if all one point set in X is open If and only if every finite set in is closed If and only if for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y}

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.21) and 3) are true, I cant tell what you mean by the last line

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[Iff for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y} \] the last statement

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\[\forall x,y\in X \ x\in U_x\text{ and } y\in U_y\] What are \(U_x, U_y\) ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2It has to be the open one.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2If every one point set was open, then we would just get the discrete topology every time, and thus the T_1 definition would be redundant.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\[\epsilon \neq \in\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Don't une `\epsilon` to mean "in" :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ux and Uy are open sets not given the open sets sir . ok. what about option 2. is it right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for tell me that sir

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let X and Y be topological spaces; let f: X\rightarrow Y be a bijection. If both f and its inverse f^{1} are continuous, then f is called _____________________ Epimorphism Homomorphism Homeomorphism Endomorpism

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2homomorphism is a group concept

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 with these properties except one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry wrong option sir

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 \[ \forall x, y \epsilon X, x \bigcap y = 0 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think they want to confuse me with that or its an error

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but a and d, i am confuse

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2these do not make sense The first one: What the hell is \(\neg\) The second one: \(U_x\in N(x)\) makes no sense, Sets are not elements in the neighborhood, they are subsets if anything. The third one: Are we to assume \(x\ne y\)? Because if they are equal then this is of course true. The last one should say \(U_x\cap U_y=\emptyset\) and this is true.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok sir, what if option A was \[If \forall x, y \epsilon\mathbb X; x \rightarrow y \] is it correct?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2can you take a screen shot?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2you latex work needs some love :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i will but its not always clear

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2here is one more thing for latest. Dont use words in the math part Like if the following: If `\(x\in X\)` then we have `\(\forall x \)`

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2keep the text outside of the inline math. or else it ill look like this \(if x\in X then \forall x\) Which was given by the following code `\(if x\in X then \forall x\)`

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2or put the text in `\text{stuff here}` Example \(\text{if }x\in X\text{ then } \forall x\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2ok close this and ask a new one if you have something else.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2tag me if you need me...

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is epsilon not same as in?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2nope. epsilon is a greek letter, in is just means ... in :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2That option makes no sense...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it din't show fully but thats how it is.ok. thank you sir

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2or it is trivially true. can you say in normal words what that option says?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Like, generally \(U_X\) means the following: An open set \(U\) that contains \(x\). But this is always true in a topological space..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but, so, that option is wrong since a set containing X can not be in the neighborhood of x

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Why cant a set containing \(x\) not be in \(N(x)\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(N(x)\subseteq N(x)\). Unless you mean \(\in\) in which case it is weird....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0N(x) means elements around X and not sets around X. i might be wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. can we try another?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2yep, yes then, if you mean \(\in\) and not \(\epsilon\), then yes it does not make sense. But it actually could This will be a little weird but you elements could actually be sets. \(\{x, \{x\}\}\) is a thing. and \(\{x\}\in \{x,\{x\}\}\) But lets move on.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2close this please, it takes to long to scroll on this pc
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