Example of a connected topological space with an open singleton set
I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:
$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.
I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.
general-topology connectedness
add a comment |
I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:
$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.
I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.
general-topology connectedness
2
How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20
1
@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36
Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52
As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24
add a comment |
I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:
$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.
I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.
general-topology connectedness
I was working on a problem that asked me to find an example of a topological space $X$ that satisfies the following:
$X$ is a connected, $X-{p}$ is disconnected and ${p}$ is open.
I have been unable to find such a space and I have almost convinced myself that there can exist no such space. Any help would be highly appreciated.
general-topology connectedness
general-topology connectedness
asked Nov 21 '18 at 20:10
Heinrich WagnerHeinrich Wagner
24919
24919
2
How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20
1
@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36
Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52
As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24
add a comment |
2
How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20
1
@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36
Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52
As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24
2
2
How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20
How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20
1
1
@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36
@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36
Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52
Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52
As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24
As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24
add a comment |
1 Answer
1
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oldest
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The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$
One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.
$X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.
add a comment |
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1 Answer
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votes
The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$
One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.
$X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.
add a comment |
The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$
One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.
$X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.
add a comment |
The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$
One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.
$X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.
The included point topology on a set will work: let $X$ be a set (say infinite) and let $p in X$ and define the included point topology $$mathcal{T}_p = {emptyset} cup {A subseteq X: p in A}$$
One easily checks this is a topology and ${p}$ is open while in $Xsetminus {p}$ all singletons are relatively open as $qneq p$ implies $${q} = {p,q} cap (X setminus {p}$$
is open in that subspace, making $Xsetminus {p}$ infinite discrete and hence disconnected.
$X$ itself is connected as any two non-empty open sets must intersect in $p$, so there can be decomposition of $X$ into two non-empty disjoint open sets.
edited Nov 22 '18 at 4:43
answered Nov 21 '18 at 22:22
Henno BrandsmaHenno Brandsma
105k347114
105k347114
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How about something like $X = { a, b, p }$ with the topology where $U$ is open if and only if $p in U$ or $U = emptyset$? (Then the subspace topology on ${ a, b }$ is the discrete topology.)
– Daniel Schepler
Nov 21 '18 at 20:20
1
@DanielSchepler You should turn that into an answer.
– freakish
Nov 21 '18 at 20:36
Bravo! That does it. Thank you for your help.
– Heinrich Wagner
Nov 21 '18 at 21:52
As soon as the space is $T_1$ we cannot have such a space, as then such an open singleton is a non-trivial closed-and-open subset of $X$.
– Henno Brandsma
Nov 21 '18 at 22:24