anonymous
  • anonymous
here
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
?

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anonymous
  • anonymous
{x: \[\frac{1}{2}\ < \] |x| < 1}
zzr0ck3r
  • zzr0ck3r
not closed
anonymous
  • anonymous
is that open?
anonymous
  • anonymous
{x:\[ \frac{1}{2} \] <\[ |x|\leqslant \]1}
zzr0ck3r
  • zzr0ck3r
yes, since it is open in R and and it equals the intersection with the subspace.
zzr0ck3r
  • zzr0ck3r
its open
anonymous
  • anonymous
\[{x: \frac{1}{2}\leqslant|x| < 1} \]
zzr0ck3r
  • zzr0ck3r
neither open nor closed
anonymous
  • anonymous
\[{x: \frac{1}{2}\leqslant|x| < 1} \]
zzr0ck3r
  • zzr0ck3r
thats the same. please use parentheses. bbl
anonymous
  • anonymous
\[{x: \frac{1}{2} < |x|\leqslant1} \]
anonymous
  • anonymous
\[{x: \frac{1}{2}\leqslant|x|\leqslant1} \]
anonymous
  • anonymous
i presume this is close
zzr0ck3r
  • zzr0ck3r
yep
zzr0ck3r
  • zzr0ck3r
closed sets in a subspace S, are of the form \(A\cap S\) where \(A\) is open in the parent topology.
zzr0ck3r
  • zzr0ck3r
ok walking out the door
anonymous
  • anonymous
and which is open in Y with same options ?
anonymous
  • anonymous
is it \[{x: \frac{1}{2}\ < |x| < 1} \]
anonymous
  • anonymous
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?
anonymous
  • anonymous
With the standard topology on R,which one of the sets in question (1) above is closed in R? with same options given above
zzr0ck3r
  • zzr0ck3r
hi
anonymous
  • anonymous
sir, i am here @zzr0ck3r
zzr0ck3r
  • zzr0ck3r
ok
anonymous
  • anonymous
so, can you help with the asked question sir?
zzr0ck3r
  • zzr0ck3r
\((-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\) 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
anonymous
  • anonymous
so which means that \[ x:1/2 <|x|<1 \] is open in the usual standard toplogy
zzr0ck3r
  • zzr0ck3r
correct
anonymous
  • anonymous
what about in the standard topology on R
anonymous
  • anonymous
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?
anonymous
  • anonymous
you here sir?
anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
hello sir, @zzr0ck3r
anonymous
  • anonymous
let me post the question in full . please don't get mad at me
anonymous
  • anonymous
With the standard topology on R,which one of the sets is open in R?
anonymous
  • anonymous
\[ {x: \frac{1}{2}\ < |x| < 1} \] \[{x: \frac{1}{2} < |x|\leqslant1} \] \[{x: \frac{1}{2}\leqslant|x| < 1} \] \[{x: \frac{1}{2}\leqslant|x|\leqslant1}\]
anonymous
  • anonymous
is the answer still the first here sir?
anonymous
  • anonymous
i guess from what you thought, A is the only open set while B,C, D are close sets
anonymous
  • anonymous
am i correct?
anonymous
  • anonymous
another question
anonymous
  • 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 wiith these properties except one.
anonymous
  • anonymous
A) if \[ \forall x, y \epsilon\mathbb X; x \neg y \]
anonymous
  • anonymous
B)\[There \exists U_x \epsilon N(x). U_y \epsilon N(y) \]
anonymous
  • anonymous
C) \[U_x \bigcap U-y = \phi \]
anonymous
  • anonymous
D)\[\forall x, y \epsilon X, x \bigcap y = 0 \]
zzr0ck3r
  • zzr0ck3r
what is the question?
anonymous
  • anonymous
should i post it in anew question?
zzr0ck3r
  • zzr0ck3r
yes

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