Question

here

Answer 1

Let R be endowed with the usual standard topology. Consider Y = [-1,1] as a subspace of R. Which one of the following sets is closed in Y

Answer 2

@zzr0ck3r

Answer 3

?

Answer 4

{x: \[\frac{1}{2}\ < \] |x| < 1}

Answer 5

not closed

Answer 6

is that open?

Answer 7

{x:\[ \frac{1}{2} \] <\[ |x|\leqslant \]1}

Answer 8

yes, since it is open in R and and it equals the intersection with the subspace.

Answer 9

its open

Answer 10

\[{x: \frac{1}{2}\leqslant|x| < 1} \]

Answer 11

neither open nor closed

Answer 12

\[{x: \frac{1}{2}\leqslant|x| < 1} \]

Answer 13

thats the same. please use parentheses. bbl

Answer 14

\[{x: \frac{1}{2} < |x|\leqslant1} \]

Answer 15

\[{x: \frac{1}{2}\leqslant|x|\leqslant1} \]

Answer 16

i presume this is close

Answer 17

yep

Answer 18

closed sets in a subspace S, are of the form \(A\cap S\) where \(A\) is open in the parent topology.

Answer 19

ok walking out the door

Answer 20

and which is open in Y with same options ?

Answer 21

is it \[{x: \frac{1}{2}\ < |x| < 1} \]

Answer 22

2 With the standard topology on R,which one of the sets in question (1) above is open in R? with same options given above?

Answer 23

With the standard topology on R,which one of the sets in question (1) above is closed in R? with same options given above

Answer 24

hi

Answer 25

sir, i am here @zzr0ck3r

Answer 26

ok

Answer 27

so, can you help with the asked question sir?

Answer 28

\((-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\) is open

\((-1, -\dfrac{1}{2}]\cup [\dfrac{1}{2}, 1)\) is not open

\([-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1]\) is not open

\([-1, -\dfrac{1}{2}]\cup [\dfrac{1}{2}, 1]\) is not open

Answer 29

so which means that \[ x:1/2 <|x|<1 \] is open in the usual standard toplogy

Answer 30

correct

Answer 31

what about in the standard topology on R

Answer 32

2 With the standard topology on R,which one of the sets in question (1) above is open in R with same options given above?

Answer 33

you here sir?

Answer 34

@zzr0ck3r

Answer 35

hello sir, @zzr0ck3r

Answer 36

let me post the question in full . please don't get mad at me

Answer 37

With the standard topology on R,which one of the sets is open in R?

Answer 38

\[ {x: \frac{1}{2}\ < |x| < 1} \]
\[{x: \frac{1}{2} < |x|\leqslant1} \]
\[{x: \frac{1}{2}\leqslant|x| < 1} \]
\[{x: \frac{1}{2}\leqslant|x|\leqslant1}\]

Answer 39

is the answer still the first here sir?

Answer 40

i guess from what you thought, A is the only open set while B,C, D are close sets

Answer 41

am i correct?

Answer 42

another question

Answer 43

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 wiith these properties except one.

Answer 44

A) if \[ \forall x, y \epsilon\mathbb X; x \neg y \]

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

what is the question?

Answer 49

should i post it in anew question?

Answer 50

yes