Why is my proof that $mathbb R$ is disconnected wrong?
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The definition of connectedness in my notes is:
A topological space $X$ is connected if there does not exist a pair of non empty subsets $U$, $V$ such that $Ucap V=emptyset$ and $Ucup V=X$.
However if I have the subsets $(-infty,0]$ and $(0,infty)$ then these are disjoint and cover $mathbb R$ and hence $mathbb R$ is disconnected.
However $mathbb R$ is clearly connected. Where have I gone wrong?
general-topology proof-verification elementary-set-theory connectedness
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
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add a comment |
$begingroup$
The definition of connectedness in my notes is:
A topological space $X$ is connected if there does not exist a pair of non empty subsets $U$, $V$ such that $Ucap V=emptyset$ and $Ucup V=X$.
However if I have the subsets $(-infty,0]$ and $(0,infty)$ then these are disjoint and cover $mathbb R$ and hence $mathbb R$ is disconnected.
However $mathbb R$ is clearly connected. Where have I gone wrong?
general-topology proof-verification elementary-set-theory connectedness
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
$endgroup$
5
$begingroup$
Open sets. You're missing the point 'open sets'.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:39
1
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Yes thank you, that would fix it
$endgroup$
– Toby Peterken
Dec 8 '18 at 11:45
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You're welcome.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:47
add a comment |
$begingroup$
The definition of connectedness in my notes is:
A topological space $X$ is connected if there does not exist a pair of non empty subsets $U$, $V$ such that $Ucap V=emptyset$ and $Ucup V=X$.
However if I have the subsets $(-infty,0]$ and $(0,infty)$ then these are disjoint and cover $mathbb R$ and hence $mathbb R$ is disconnected.
However $mathbb R$ is clearly connected. Where have I gone wrong?
general-topology proof-verification elementary-set-theory connectedness
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
$endgroup$
The definition of connectedness in my notes is:
A topological space $X$ is connected if there does not exist a pair of non empty subsets $U$, $V$ such that $Ucap V=emptyset$ and $Ucup V=X$.
However if I have the subsets $(-infty,0]$ and $(0,infty)$ then these are disjoint and cover $mathbb R$ and hence $mathbb R$ is disconnected.
However $mathbb R$ is clearly connected. Where have I gone wrong?
general-topology proof-verification elementary-set-theory connectedness
general-topology proof-verification elementary-set-theory connectedness
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
asked Dec 8 '18 at 11:34
Toby PeterkenToby Peterken
1496
1496
5
$begingroup$
Open sets. You're missing the point 'open sets'.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:39
1
$begingroup$
Yes thank you, that would fix it
$endgroup$
– Toby Peterken
Dec 8 '18 at 11:45
$begingroup$
You're welcome.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:47
add a comment |
5
$begingroup$
Open sets. You're missing the point 'open sets'.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:39
1
$begingroup$
Yes thank you, that would fix it
$endgroup$
– Toby Peterken
Dec 8 '18 at 11:45
$begingroup$
You're welcome.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:47
5
5
$begingroup$
Open sets. You're missing the point 'open sets'.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:39
$begingroup$
Open sets. You're missing the point 'open sets'.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:39
1
1
$begingroup$
Yes thank you, that would fix it
$endgroup$
– Toby Peterken
Dec 8 '18 at 11:45
$begingroup$
Yes thank you, that would fix it
$endgroup$
– Toby Peterken
Dec 8 '18 at 11:45
$begingroup$
You're welcome.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:47
$begingroup$
You're welcome.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
With your definition, every space $X$ with at least two points would be disconnected: just take a point $xin X$ and consider $X={x}cup(Xsetminus{x})$.
The definition requires $U$ and $V$ to be disjoint nonempty open sets such that $Ucup V=X$.
The set $(-infty,0]$ is not open.
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
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add a comment |
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The subsets that you take are wrong because $(-infty ,0]$ contains a accumulation point of $(0,infty)$.
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
With your definition, every space $X$ with at least two points would be disconnected: just take a point $xin X$ and consider $X={x}cup(Xsetminus{x})$.
The definition requires $U$ and $V$ to be disjoint nonempty open sets such that $Ucup V=X$.
The set $(-infty,0]$ is not open.
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
With your definition, every space $X$ with at least two points would be disconnected: just take a point $xin X$ and consider $X={x}cup(Xsetminus{x})$.
The definition requires $U$ and $V$ to be disjoint nonempty open sets such that $Ucup V=X$.
The set $(-infty,0]$ is not open.
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
With your definition, every space $X$ with at least two points would be disconnected: just take a point $xin X$ and consider $X={x}cup(Xsetminus{x})$.
The definition requires $U$ and $V$ to be disjoint nonempty open sets such that $Ucup V=X$.
The set $(-infty,0]$ is not open.
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
$endgroup$
With your definition, every space $X$ with at least two points would be disconnected: just take a point $xin X$ and consider $X={x}cup(Xsetminus{x})$.
The definition requires $U$ and $V$ to be disjoint nonempty open sets such that $Ucup V=X$.
The set $(-infty,0]$ is not open.
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
answered Dec 8 '18 at 11:55
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
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The subsets that you take are wrong because $(-infty ,0]$ contains a accumulation point of $(0,infty)$.
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
$endgroup$
add a comment |
$begingroup$
The subsets that you take are wrong because $(-infty ,0]$ contains a accumulation point of $(0,infty)$.
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
$endgroup$
add a comment |
$begingroup$
The subsets that you take are wrong because $(-infty ,0]$ contains a accumulation point of $(0,infty)$.
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
$endgroup$
The subsets that you take are wrong because $(-infty ,0]$ contains a accumulation point of $(0,infty)$.
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
answered Dec 8 '18 at 11:52
Fernando cañizaresFernando cañizares
11
11
add a comment |
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5
$begingroup$
Open sets. You're missing the point 'open sets'.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:39
1
$begingroup$
Yes thank you, that would fix it
$endgroup$
– Toby Peterken
Dec 8 '18 at 11:45
$begingroup$
You're welcome.
$endgroup$
– Anik Bhowmick
Dec 8 '18 at 11:47