## 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

1. anonymous

i think frechet space

2. anonymous

@zzr0ck3r

3. zzr0ck3r

correct

4. 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}

5. anonymous

@zzr0ck3r

6. zzr0ck3r

7. zzr0ck3r

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

8. anonymous

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

9. zzr0ck3r

$\forall x,y\in X \ x\in U_x\text{ and } y\in U_y$ What are $$U_x, U_y$$ ?

10. zzr0ck3r

It has to be the open one.

11. 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.

12. zzr0ck3r

What open sets?

13. zzr0ck3r

$\epsilon \neq \in$

14. zzr0ck3r

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

15. anonymous

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

16. anonymous

thanks for tell me that sir

17. zzr0ck3r

2) is not true

18. zzr0ck3r

unless it is finite.

19. zzr0ck3r

lol

20. 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

21. zzr0ck3r

homeomorphism

22. zzr0ck3r

homomorphism is a group concept

23. anonymous

sorry C

24. zzr0ck3r

correct

25. 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.

26. anonymous

sorry wrong option sir

27. anonymous

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

28. anonymous

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

29. anonymous

c $\forall x, y \epsilon X, x \bigcap y = 0$

30. zzr0ck3r

what is $$\neg$$?

31. anonymous

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

32. anonymous

$U_x \bigcap U-y = \phi$

33. anonymous

to me C is true

34. anonymous

B is also true

35. anonymous

but a and d, i am confuse

36. 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.

37. anonymous

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

38. zzr0ck3r

can you take a screen shot?

39. zzr0ck3r

you latex work needs some love :)

40. anonymous

lol

41. anonymous

i will but its not always clear

42. 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 $$\forall x$$

43. 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 $$if x\in X then \forall x$$

44. zzr0ck3r

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

45. zzr0ck3r

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

46. zzr0ck3r

tag me if you need me...

47. anonymous

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

48. anonymous

@zzr0ck3r

49. anonymous

is epsilon not same as in?

50. zzr0ck3r

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

51. zzr0ck3r

That option makes no sense...

52. anonymous

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

53. zzr0ck3r

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

54. zzr0ck3r

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

55. anonymous

hmmm

56. zzr0ck3r

Are you on windows?

57. anonymous

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

58. anonymous

no sir.

59. zzr0ck3r

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

60. zzr0ck3r

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

61. anonymous

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

62. anonymous

ok. can we try another?

63. 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.

64. zzr0ck3r

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