Question

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

Answer 1

i think frechet space

Answer 2

@zzr0ck3r

Answer 3

correct

Answer 4

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}

Answer 5

@zzr0ck3r

Answer 6

chill, reading

Answer 7

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

Answer 8

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

Answer 9

\[\forall x,y\in X \ x\in U_x\text{ and } y\in U_y\]

What are \(U_x, U_y\) ?

Answer 10

It has to be the open one.

Answer 11

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.

Answer 12

What open sets?

Answer 13

\[\epsilon \neq \in\]

Answer 14

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

Answer 15

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

Answer 16

thanks for tell me that sir

Answer 17

2) is not true

Answer 18

unless it is finite.

Answer 19

lol

Answer 20

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

Answer 21

homeomorphism

Answer 22

homomorphism is a group concept

Answer 23

sorry C

Answer 24

correct

Answer 25

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.

Answer 26

sorry wrong option sir

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

what is \(\neg\)?

Answer 31

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

Answer 32

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

Answer 33

to me C is true

Answer 34

B is also true

Answer 35

but a and d, i am confuse

Answer 36

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.

Answer 37

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

Answer 38

can you take a screen shot?

Answer 39

you latex work needs some love :)

Answer 40

lol

Answer 41

i will but its not always clear

Answer 42

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 `\(\forall x \)`

Answer 43

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
`\(if x\in X then \forall x\)`

Answer 44

or put the text in `\text{stuff here}`

Example

\(\text{if }x\in X\text{ then } \forall x\)

Answer 45

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

Answer 46

tag me if you need me...

Answer 47

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

Answer 48

@zzr0ck3r

Answer 49

is epsilon not same as in?

Answer 50

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

Answer 51

That option makes no sense...

Answer 52

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

Answer 53

or it is trivially true.

can you say in normal words what that option says?

Answer 54

Like, generally \(U_X\) means the following:

An open set \(U\) that contains \(x\). But this is always true in a topological space..

Answer 55

hmmm

Answer 56

Are you on windows?

Answer 57

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

Answer 58

no sir.

Answer 59

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

Answer 60

\(N(x)\subseteq N(x)\).

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

Answer 61

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

Answer 62

ok. can we try another?

Answer 63

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.

Answer 64

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