What are the accumulation points of ( $pi,2pi$ ] $subsetmathbb{R}$?
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I was just curious as to what the accumulation points of ( $pi,2pi$ ] $subsetmathbb{R}$ are. I had gotten $pi,2pi$,1. But the other answers are 3 and 5.
real-analysis
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
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add a comment |
$begingroup$
I was just curious as to what the accumulation points of ( $pi,2pi$ ] $subsetmathbb{R}$ are. I had gotten $pi,2pi$,1. But the other answers are 3 and 5.
real-analysis
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
$endgroup$
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can you include some workings how do you obtain those partial results?
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– Siong Thye Goh
Nov 26 '18 at 3:05
add a comment |
$begingroup$
I was just curious as to what the accumulation points of ( $pi,2pi$ ] $subsetmathbb{R}$ are. I had gotten $pi,2pi$,1. But the other answers are 3 and 5.
real-analysis
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
$endgroup$
I was just curious as to what the accumulation points of ( $pi,2pi$ ] $subsetmathbb{R}$ are. I had gotten $pi,2pi$,1. But the other answers are 3 and 5.
real-analysis
real-analysis
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
asked Nov 26 '18 at 2:57
Karl TeschKarl Tesch
1
1
$begingroup$
can you include some workings how do you obtain those partial results?
$endgroup$
– Siong Thye Goh
Nov 26 '18 at 3:05
add a comment |
$begingroup$
can you include some workings how do you obtain those partial results?
$endgroup$
– Siong Thye Goh
Nov 26 '18 at 3:05
$begingroup$
can you include some workings how do you obtain those partial results?
$endgroup$
– Siong Thye Goh
Nov 26 '18 at 3:05
$begingroup$
can you include some workings how do you obtain those partial results?
$endgroup$
– Siong Thye Goh
Nov 26 '18 at 3:05
add a comment |
1 Answer
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Every point in $[pi, 2 pi]$ is an accumulation point of $(pi, 2pi]$ since any neighbourhood of any point of in $[pi, 2 pi]$ contains a point of $(pi, 2pi]$.
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
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$begingroup$
Every point in $[pi, 2 pi]$ is an accumulation point of $(pi, 2pi]$ since any neighbourhood of any point of in $[pi, 2 pi]$ contains a point of $(pi, 2pi]$.
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
$endgroup$
add a comment |
$begingroup$
Every point in $[pi, 2 pi]$ is an accumulation point of $(pi, 2pi]$ since any neighbourhood of any point of in $[pi, 2 pi]$ contains a point of $(pi, 2pi]$.
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
$endgroup$
add a comment |
$begingroup$
Every point in $[pi, 2 pi]$ is an accumulation point of $(pi, 2pi]$ since any neighbourhood of any point of in $[pi, 2 pi]$ contains a point of $(pi, 2pi]$.
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
$endgroup$
Every point in $[pi, 2 pi]$ is an accumulation point of $(pi, 2pi]$ since any neighbourhood of any point of in $[pi, 2 pi]$ contains a point of $(pi, 2pi]$.
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
answered Nov 26 '18 at 3:07
SheafsSheafs
1517
1517
add a comment |
add a comment |
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$begingroup$
can you include some workings how do you obtain those partial results?
$endgroup$
– Siong Thye Goh
Nov 26 '18 at 3:05