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

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- anonymous

i think frechet space

- anonymous

@zzr0ck3r

- zzr0ck3r

correct

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## More answers

- anonymous

One 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}

- anonymous

@zzr0ck3r

- zzr0ck3r

chill, reading

- zzr0ck3r

1) and 3) are true, I cant tell what you mean by the last line

- anonymous

\[Iff for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y} \] the last statement

- zzr0ck3r

\[\forall x,y\in X \ x\in U_x\text{ and } y\in U_y\]
What are \(U_x, U_y\) ?

- zzr0ck3r

It has to be the open one.

- zzr0ck3r

If 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

What open sets?

- zzr0ck3r

\[\epsilon \neq \in\]

- zzr0ck3r

Don't une `\epsilon` to mean "in" :)

- anonymous

Ux and Uy are open sets not given the open sets sir .
ok. what about option 2. is it right?

- anonymous

thanks for tell me that sir

- zzr0ck3r

2) is not true

- zzr0ck3r

unless it is finite.

- zzr0ck3r

lol

- anonymous

Let 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

homeomorphism

- zzr0ck3r

homomorphism is a group concept

- anonymous

sorry C

- zzr0ck3r

correct

- anonymous

A 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

sorry wrong option sir

- anonymous

a \[If \forall x, y \epsilon\mathbb X; x \neg y \]

- anonymous

b \[There \exists U_x \epsilon N(x). U_y \epsilon N(y) \]

- anonymous

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

- zzr0ck3r

what is \(\neg\)?

- anonymous

i think they want to confuse me with that or its an error

- anonymous

\[U_x \bigcap U-y = \phi \]

- anonymous

to me C is true

- anonymous

B is also true

- anonymous

but a and d, i am confuse

- zzr0ck3r

these 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

ok sir, what if option A was
\[If \forall x, y \epsilon\mathbb X; x \rightarrow y \] is it correct?

- zzr0ck3r

can you take a screen shot?

- zzr0ck3r

you latex work needs some love :)

- anonymous

lol

- anonymous

i will but its not always clear

- zzr0ck3r

here 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

keep 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

or put the text in `\text{stuff here}`
Example
\(\text{if }x\in X\text{ then } \forall x\)

- zzr0ck3r

ok close this and ask a new one if you have something else.

- zzr0ck3r

tag me if you need me...

- anonymous

There exists U_x \epsilon N(x). U_y \epsilon N(y) here is how option B is

- anonymous

@zzr0ck3r

- anonymous

is epsilon not same as in?

- zzr0ck3r

nope. epsilon is a greek letter, in is just means ... in :)

- zzr0ck3r

That option makes no sense...

- anonymous

it din't show fully but thats how it is.ok. thank you sir

- zzr0ck3r

or it is trivially true.
can you say in normal words what that option says?

- zzr0ck3r

Like, generally \(U_X\) means the following:
An open set \(U\) that contains \(x\). But this is always true in a topological space..

- anonymous

hmmm

- zzr0ck3r

Are you on windows?

- anonymous

but, so, that option is wrong since a set containing X can not be in the neighborhood of x

- anonymous

no sir.

- zzr0ck3r

Why cant a set containing \(x\) not be in \(N(x)\)?

- zzr0ck3r

\(N(x)\subseteq N(x)\).
Unless you mean \(\in\) in which case it is weird....

- anonymous

N(x) means elements around X and not sets around X. i might be wrong

- anonymous

ok. can we try another?

- zzr0ck3r

yep, 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

close this please, it takes to long to scroll on this pc

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