Equality of two topological spaces
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Suppose $(X, T_1), (X, T_2)$ be two topological spaces. It is given that a sequence ${x_n}$ is convergent in $T_1$ if and only if it is convergent in $T_2.$ Does this imply $T_1= T_2;$
?
If not, I can't find any counterexamples.
sequences-and-series general-topology convergence
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
asked Jan 7 at 17:31
KkDKkD
1
$endgroup$
add a comment |
$begingroup$
Suppose $(X, T_1), (X, T_2)$ be two topological spaces. It is given that a sequence ${x_n}$ is convergent in $T_1$ if and only if it is convergent in $T_2.$ Does this imply $T_1= T_2;$
?
If not, I can't find any counterexamples.
sequences-and-series general-topology convergence
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
asked Jan 7 at 17:31
KkDKkD
1
$endgroup$
$begingroup$
Do sequences converge vacuously in a topology which isn't metrizable? Or are you allowing sequences defined by nets, etc?
$endgroup$
– Chickenmancer
Jan 7 at 17:36
1
$begingroup$
A sequence {xn} in (X,T) is convergent to some x in X if for any open U containing x, there is some m in natural numbers such that xn is in U for all n>m.
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– KkD
Jan 8 at 3:31
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How about $X=Bbb R$, $T_1$ is the set of intervals $(-s, s)$, and $T_2$ is the subset of $T_1$ where $s=frac1k$ for some integer $k$.
$endgroup$
– MJD
Jan 8 at 5:07
add a comment |
$begingroup$
Suppose $(X, T_1), (X, T_2)$ be two topological spaces. It is given that a sequence ${x_n}$ is convergent in $T_1$ if and only if it is convergent in $T_2.$ Does this imply $T_1= T_2;$
?
If not, I can't find any counterexamples.
sequences-and-series general-topology convergence
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
asked Jan 7 at 17:31
KkDKkD
1
$endgroup$
Suppose $(X, T_1), (X, T_2)$ be two topological spaces. It is given that a sequence ${x_n}$ is convergent in $T_1$ if and only if it is convergent in $T_2.$ Does this imply $T_1= T_2;$
?
If not, I can't find any counterexamples.
sequences-and-series general-topology convergence
sequences-and-series general-topology convergence
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
asked Jan 7 at 17:31
KkDKkD
1
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
asked Jan 7 at 17:31
KkDKkD
1
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
edited Jan 7 at 22:00
José Carlos Santos
171k23132240
171k23132240
asked Jan 7 at 17:31
KkDKkD
1
asked Jan 7 at 17:31
KkDKkD
1
asked Jan 7 at 17:31
KkDKkD
1
1
$begingroup$
Do sequences converge vacuously in a topology which isn't metrizable? Or are you allowing sequences defined by nets, etc?
$endgroup$
– Chickenmancer
Jan 7 at 17:36
1
$begingroup$
A sequence {xn} in (X,T) is convergent to some x in X if for any open U containing x, there is some m in natural numbers such that xn is in U for all n>m.
$endgroup$
– KkD
Jan 8 at 3:31
$begingroup$
How about $X=Bbb R$, $T_1$ is the set of intervals $(-s, s)$, and $T_2$ is the subset of $T_1$ where $s=frac1k$ for some integer $k$.
$endgroup$
– MJD
Jan 8 at 5:07
add a comment |
$begingroup$
Do sequences converge vacuously in a topology which isn't metrizable? Or are you allowing sequences defined by nets, etc?
$endgroup$
– Chickenmancer
Jan 7 at 17:36
1
$begingroup$
A sequence {xn} in (X,T) is convergent to some x in X if for any open U containing x, there is some m in natural numbers such that xn is in U for all n>m.
$endgroup$
– KkD
Jan 8 at 3:31
$begingroup$
How about $X=Bbb R$, $T_1$ is the set of intervals $(-s, s)$, and $T_2$ is the subset of $T_1$ where $s=frac1k$ for some integer $k$.
$endgroup$
– MJD
Jan 8 at 5:07
$begingroup$
Do sequences converge vacuously in a topology which isn't metrizable? Or are you allowing sequences defined by nets, etc?
$endgroup$
– Chickenmancer
Jan 7 at 17:36
$begingroup$
Do sequences converge vacuously in a topology which isn't metrizable? Or are you allowing sequences defined by nets, etc?
$endgroup$
– Chickenmancer
Jan 7 at 17:36
1
1
$begingroup$
A sequence {xn} in (X,T) is convergent to some x in X if for any open U containing x, there is some m in natural numbers such that xn is in U for all n>m.
$endgroup$
– KkD
Jan 8 at 3:31
$begingroup$
A sequence {xn} in (X,T) is convergent to some x in X if for any open U containing x, there is some m in natural numbers such that xn is in U for all n>m.
$endgroup$
– KkD
Jan 8 at 3:31
$begingroup$
How about $X=Bbb R$, $T_1$ is the set of intervals $(-s, s)$, and $T_2$ is the subset of $T_1$ where $s=frac1k$ for some integer $k$.
$endgroup$
– MJD
Jan 8 at 5:07
$begingroup$
How about $X=Bbb R$, $T_1$ is the set of intervals $(-s, s)$, and $T_2$ is the subset of $T_1$ where $s=frac1k$ for some integer $k$.
$endgroup$
– MJD
Jan 8 at 5:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is negative. Consider two topologies $tau_1$ and $tau_2$ in $mathbb R$:
$tau_1$ is the discrete topology;
$tau_2$ is the topology for which a set $Sneqmathbb R$ is closed if and only if $S$ is finite or countable.
Obviously, $tau_1neqtau_2$. Now, prove that the convergent sequences are the same in $(mathbb{R},tau_1)$ and in $(mathbb{R},tau_2)$.
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
consider $omega_1+1$ with the order topology $tau$, and the topology generated by $taucup{{omega_1}}$.
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The answer is negative. Consider two topologies $tau_1$ and $tau_2$ in $mathbb R$:
$tau_1$ is the discrete topology;
$tau_2$ is the topology for which a set $Sneqmathbb R$ is closed if and only if $S$ is finite or countable.
Obviously, $tau_1neqtau_2$. Now, prove that the convergent sequences are the same in $(mathbb{R},tau_1)$ and in $(mathbb{R},tau_2)$.
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
The answer is negative. Consider two topologies $tau_1$ and $tau_2$ in $mathbb R$:
$tau_1$ is the discrete topology;
$tau_2$ is the topology for which a set $Sneqmathbb R$ is closed if and only if $S$ is finite or countable.
Obviously, $tau_1neqtau_2$. Now, prove that the convergent sequences are the same in $(mathbb{R},tau_1)$ and in $(mathbb{R},tau_2)$.
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
The answer is negative. Consider two topologies $tau_1$ and $tau_2$ in $mathbb R$:
$tau_1$ is the discrete topology;
$tau_2$ is the topology for which a set $Sneqmathbb R$ is closed if and only if $S$ is finite or countable.
Obviously, $tau_1neqtau_2$. Now, prove that the convergent sequences are the same in $(mathbb{R},tau_1)$ and in $(mathbb{R},tau_2)$.
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
The answer is negative. Consider two topologies $tau_1$ and $tau_2$ in $mathbb R$:
$tau_1$ is the discrete topology;
$tau_2$ is the topology for which a set $Sneqmathbb R$ is closed if and only if $S$ is finite or countable.
Obviously, $tau_1neqtau_2$. Now, prove that the convergent sequences are the same in $(mathbb{R},tau_1)$ and in $(mathbb{R},tau_2)$.
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 7 at 17:38
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
consider $omega_1+1$ with the order topology $tau$, and the topology generated by $taucup{{omega_1}}$.
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
$endgroup$
add a comment |
$begingroup$
consider $omega_1+1$ with the order topology $tau$, and the topology generated by $taucup{{omega_1}}$.
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
$endgroup$
add a comment |
$begingroup$
consider $omega_1+1$ with the order topology $tau$, and the topology generated by $taucup{{omega_1}}$.
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
$endgroup$
consider $omega_1+1$ with the order topology $tau$, and the topology generated by $taucup{{omega_1}}$.
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
answered Jan 8 at 4:47
Jorge Cruz ChapitalJorge Cruz Chapital
1
1
add a comment |
add a comment |
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$begingroup$
Do sequences converge vacuously in a topology which isn't metrizable? Or are you allowing sequences defined by nets, etc?
$endgroup$
– Chickenmancer
Jan 7 at 17:36
1
$begingroup$
A sequence {xn} in (X,T) is convergent to some x in X if for any open U containing x, there is some m in natural numbers such that xn is in U for all n>m.
$endgroup$
– KkD
Jan 8 at 3:31
$begingroup$
How about $X=Bbb R$, $T_1$ is the set of intervals $(-s, s)$, and $T_2$ is the subset of $T_1$ where $s=frac1k$ for some integer $k$.
$endgroup$
– MJD
Jan 8 at 5:07