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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
?

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

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