Proof that subspace topology is independent of 'parent' space
$begingroup$
In my topology textbook (Bert Mendelson's) it is stated that if $C$ is a subspace of two distinct larger spaces $X$ and $Y$, then the relative topology of $C$ is the same whether we regard it as a subspace of $X$ or $Y$.
In an attempt to prove this, regard $C$ as a subspace of $X$ and suppose $S$ is an open subset of $C$. Then $$S = S' cap C$$ for some open subset $S'$ of $X$. Now I'm not sure how to continue; proving that $S' cap Y$ is an open subset of $Y$ would do the job, since then $$S' cap Y cap C = S$$ is in the topology of $C$ regarded as a subspace of $Y$, but I don't see a way to prove this.
Can anyone help me with this proof? Can $S' cap Y$ be shown to be open in $Y$ or should I take a completely different approach?
Thanks in advance!
general-topology
edited Dec 21 '18 at 23:08
egreg
183k1486205
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
$endgroup$
|
show 10 more comments
$begingroup$
In my topology textbook (Bert Mendelson's) it is stated that if $C$ is a subspace of two distinct larger spaces $X$ and $Y$, then the relative topology of $C$ is the same whether we regard it as a subspace of $X$ or $Y$.
In an attempt to prove this, regard $C$ as a subspace of $X$ and suppose $S$ is an open subset of $C$. Then $$S = S' cap C$$ for some open subset $S'$ of $X$. Now I'm not sure how to continue; proving that $S' cap Y$ is an open subset of $Y$ would do the job, since then $$S' cap Y cap C = S$$ is in the topology of $C$ regarded as a subspace of $Y$, but I don't see a way to prove this.
Can anyone help me with this proof? Can $S' cap Y$ be shown to be open in $Y$ or should I take a completely different approach?
Thanks in advance!
general-topology
edited Dec 21 '18 at 23:08
egreg
183k1486205
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
$endgroup$
$begingroup$
By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition
$endgroup$
– Ben W
Dec 21 '18 at 15:32
2
$begingroup$
What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 15:33
$begingroup$
I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done.
$endgroup$
– YuiTo Cheng
Dec 21 '18 at 15:33
$begingroup$
@HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..."
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:36
$begingroup$
So the one is not necessarily a subspace of the other.
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:37
|
show 10 more comments
$begingroup$
In my topology textbook (Bert Mendelson's) it is stated that if $C$ is a subspace of two distinct larger spaces $X$ and $Y$, then the relative topology of $C$ is the same whether we regard it as a subspace of $X$ or $Y$.
In an attempt to prove this, regard $C$ as a subspace of $X$ and suppose $S$ is an open subset of $C$. Then $$S = S' cap C$$ for some open subset $S'$ of $X$. Now I'm not sure how to continue; proving that $S' cap Y$ is an open subset of $Y$ would do the job, since then $$S' cap Y cap C = S$$ is in the topology of $C$ regarded as a subspace of $Y$, but I don't see a way to prove this.
Can anyone help me with this proof? Can $S' cap Y$ be shown to be open in $Y$ or should I take a completely different approach?
Thanks in advance!
general-topology
edited Dec 21 '18 at 23:08
egreg
183k1486205
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
$endgroup$
In my topology textbook (Bert Mendelson's) it is stated that if $C$ is a subspace of two distinct larger spaces $X$ and $Y$, then the relative topology of $C$ is the same whether we regard it as a subspace of $X$ or $Y$.
In an attempt to prove this, regard $C$ as a subspace of $X$ and suppose $S$ is an open subset of $C$. Then $$S = S' cap C$$ for some open subset $S'$ of $X$. Now I'm not sure how to continue; proving that $S' cap Y$ is an open subset of $Y$ would do the job, since then $$S' cap Y cap C = S$$ is in the topology of $C$ regarded as a subspace of $Y$, but I don't see a way to prove this.
Can anyone help me with this proof? Can $S' cap Y$ be shown to be open in $Y$ or should I take a completely different approach?
Thanks in advance!
general-topology
general-topology
edited Dec 21 '18 at 23:08
egreg
183k1486205
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
edited Dec 21 '18 at 23:08
egreg
183k1486205
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
edited Dec 21 '18 at 23:08
egreg
183k1486205
edited Dec 21 '18 at 23:08
egreg
183k1486205
edited Dec 21 '18 at 23:08
egreg
183k1486205
183k1486205
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
asked Dec 21 '18 at 15:25
Steven WagterSteven Wagter
1569
1569
$begingroup$
By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition
$endgroup$
– Ben W
Dec 21 '18 at 15:32
2
$begingroup$
What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 15:33
$begingroup$
I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done.
$endgroup$
– YuiTo Cheng
Dec 21 '18 at 15:33
$begingroup$
@HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..."
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:36
$begingroup$
So the one is not necessarily a subspace of the other.
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:37
|
show 10 more comments
$begingroup$
By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition
$endgroup$
– Ben W
Dec 21 '18 at 15:32
2
$begingroup$
What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 15:33
$begingroup$
I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done.
$endgroup$
– YuiTo Cheng
Dec 21 '18 at 15:33
$begingroup$
@HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..."
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:36
$begingroup$
So the one is not necessarily a subspace of the other.
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:37
$begingroup$
By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition
$endgroup$
– Ben W
Dec 21 '18 at 15:32
$begingroup$
By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition
$endgroup$
– Ben W
Dec 21 '18 at 15:32
2
2
$begingroup$
What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 15:33
$begingroup$
What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 15:33
$begingroup$
I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done.
$endgroup$
– YuiTo Cheng
Dec 21 '18 at 15:33
$begingroup$
I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done.
$endgroup$
– YuiTo Cheng
Dec 21 '18 at 15:33
$begingroup$
@HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..."
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:36
$begingroup$
@HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..."
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:36
$begingroup$
So the one is not necessarily a subspace of the other.
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:37
$begingroup$
So the one is not necessarily a subspace of the other.
$endgroup$
– Steven Wagter
Dec 21 '18 at 15:37
|
show 10 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Following the discussion in the comments I'll assume the following situation:
We have a superspace $Z$ such that $X,Y subseteq Z$ have the subspace topology w.r.t. $Z$ and $C subseteq X cap Y$.
Now $C$ can inherit the subspace topology of $X$ or of $Y$, but this does not matter, because in the end it's just the subspace topology from $Z$.
If $S subseteq C$ is open "via $X$" then $S = S_X cap C$ with $S_X$ open in $X$, so really, $S_X = O cap X$ where $O$ is open in $Z$. Hence $S = (O cap X) cap C = O cap C$ (as $X cap C = C$) and so $S$ is open as a subspace of $Z$. Moreover, $S = O cap C = O cap (C cap Y) = (O cap Y) cap C$ where $O cap Y$ is open in $Y$ (by definition of the subspace topology on $Y$) and so $S$ is also open "via $Y$". This argument is entirely symmetrical, so indeed it does not matter via which subspace $X$ or $Y$ we endow $C$ with a subspace topology. The common superspace enforces the consistency, as it were.
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
We almost always talk of a top. space $X$ when we really mean a pair $(X,T_X)$ where $T_X$ is a certain kind of collection of subsets of $X$, called a topology on $X$, and commonly called the collection of open sets. $T_X$ is not determined by $X$ alone, unless $X$ has at most one member. Suppose $T_X$ is a topology on $X$ and $T_Y$ is a topology on $Y,$ and that $Xsupset Csubset Y.$ There may be many possible topologies on the set $C$. Suppose $T_C$ is a topology on $C$ such that $(C,T_C)$ is a sub-space of $(X,T_X)$ and also of $(Y,T_Y)$. By the definition of a sub-space topology, this means that $${xcap C:xin T_X}=T_C={ycap C: yin T_Y}.$$
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Following the discussion in the comments I'll assume the following situation:
We have a superspace $Z$ such that $X,Y subseteq Z$ have the subspace topology w.r.t. $Z$ and $C subseteq X cap Y$.
Now $C$ can inherit the subspace topology of $X$ or of $Y$, but this does not matter, because in the end it's just the subspace topology from $Z$.
If $S subseteq C$ is open "via $X$" then $S = S_X cap C$ with $S_X$ open in $X$, so really, $S_X = O cap X$ where $O$ is open in $Z$. Hence $S = (O cap X) cap C = O cap C$ (as $X cap C = C$) and so $S$ is open as a subspace of $Z$. Moreover, $S = O cap C = O cap (C cap Y) = (O cap Y) cap C$ where $O cap Y$ is open in $Y$ (by definition of the subspace topology on $Y$) and so $S$ is also open "via $Y$". This argument is entirely symmetrical, so indeed it does not matter via which subspace $X$ or $Y$ we endow $C$ with a subspace topology. The common superspace enforces the consistency, as it were.
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
Following the discussion in the comments I'll assume the following situation:
We have a superspace $Z$ such that $X,Y subseteq Z$ have the subspace topology w.r.t. $Z$ and $C subseteq X cap Y$.
Now $C$ can inherit the subspace topology of $X$ or of $Y$, but this does not matter, because in the end it's just the subspace topology from $Z$.
If $S subseteq C$ is open "via $X$" then $S = S_X cap C$ with $S_X$ open in $X$, so really, $S_X = O cap X$ where $O$ is open in $Z$. Hence $S = (O cap X) cap C = O cap C$ (as $X cap C = C$) and so $S$ is open as a subspace of $Z$. Moreover, $S = O cap C = O cap (C cap Y) = (O cap Y) cap C$ where $O cap Y$ is open in $Y$ (by definition of the subspace topology on $Y$) and so $S$ is also open "via $Y$". This argument is entirely symmetrical, so indeed it does not matter via which subspace $X$ or $Y$ we endow $C$ with a subspace topology. The common superspace enforces the consistency, as it were.
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
Following the discussion in the comments I'll assume the following situation:
We have a superspace $Z$ such that $X,Y subseteq Z$ have the subspace topology w.r.t. $Z$ and $C subseteq X cap Y$.
Now $C$ can inherit the subspace topology of $X$ or of $Y$, but this does not matter, because in the end it's just the subspace topology from $Z$.
If $S subseteq C$ is open "via $X$" then $S = S_X cap C$ with $S_X$ open in $X$, so really, $S_X = O cap X$ where $O$ is open in $Z$. Hence $S = (O cap X) cap C = O cap C$ (as $X cap C = C$) and so $S$ is open as a subspace of $Z$. Moreover, $S = O cap C = O cap (C cap Y) = (O cap Y) cap C$ where $O cap Y$ is open in $Y$ (by definition of the subspace topology on $Y$) and so $S$ is also open "via $Y$". This argument is entirely symmetrical, so indeed it does not matter via which subspace $X$ or $Y$ we endow $C$ with a subspace topology. The common superspace enforces the consistency, as it were.
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
Following the discussion in the comments I'll assume the following situation:
We have a superspace $Z$ such that $X,Y subseteq Z$ have the subspace topology w.r.t. $Z$ and $C subseteq X cap Y$.
Now $C$ can inherit the subspace topology of $X$ or of $Y$, but this does not matter, because in the end it's just the subspace topology from $Z$.
If $S subseteq C$ is open "via $X$" then $S = S_X cap C$ with $S_X$ open in $X$, so really, $S_X = O cap X$ where $O$ is open in $Z$. Hence $S = (O cap X) cap C = O cap C$ (as $X cap C = C$) and so $S$ is open as a subspace of $Z$. Moreover, $S = O cap C = O cap (C cap Y) = (O cap Y) cap C$ where $O cap Y$ is open in $Y$ (by definition of the subspace topology on $Y$) and so $S$ is also open "via $Y$". This argument is entirely symmetrical, so indeed it does not matter via which subspace $X$ or $Y$ we endow $C$ with a subspace topology. The common superspace enforces the consistency, as it were.
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 21 '18 at 23:04
Henno BrandsmaHenno Brandsma
111k348118
111k348118
add a comment |
add a comment |
$begingroup$
We almost always talk of a top. space $X$ when we really mean a pair $(X,T_X)$ where $T_X$ is a certain kind of collection of subsets of $X$, called a topology on $X$, and commonly called the collection of open sets. $T_X$ is not determined by $X$ alone, unless $X$ has at most one member. Suppose $T_X$ is a topology on $X$ and $T_Y$ is a topology on $Y,$ and that $Xsupset Csubset Y.$ There may be many possible topologies on the set $C$. Suppose $T_C$ is a topology on $C$ such that $(C,T_C)$ is a sub-space of $(X,T_X)$ and also of $(Y,T_Y)$. By the definition of a sub-space topology, this means that $${xcap C:xin T_X}=T_C={ycap C: yin T_Y}.$$
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
We almost always talk of a top. space $X$ when we really mean a pair $(X,T_X)$ where $T_X$ is a certain kind of collection of subsets of $X$, called a topology on $X$, and commonly called the collection of open sets. $T_X$ is not determined by $X$ alone, unless $X$ has at most one member. Suppose $T_X$ is a topology on $X$ and $T_Y$ is a topology on $Y,$ and that $Xsupset Csubset Y.$ There may be many possible topologies on the set $C$. Suppose $T_C$ is a topology on $C$ such that $(C,T_C)$ is a sub-space of $(X,T_X)$ and also of $(Y,T_Y)$. By the definition of a sub-space topology, this means that $${xcap C:xin T_X}=T_C={ycap C: yin T_Y}.$$
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
We almost always talk of a top. space $X$ when we really mean a pair $(X,T_X)$ where $T_X$ is a certain kind of collection of subsets of $X$, called a topology on $X$, and commonly called the collection of open sets. $T_X$ is not determined by $X$ alone, unless $X$ has at most one member. Suppose $T_X$ is a topology on $X$ and $T_Y$ is a topology on $Y,$ and that $Xsupset Csubset Y.$ There may be many possible topologies on the set $C$. Suppose $T_C$ is a topology on $C$ such that $(C,T_C)$ is a sub-space of $(X,T_X)$ and also of $(Y,T_Y)$. By the definition of a sub-space topology, this means that $${xcap C:xin T_X}=T_C={ycap C: yin T_Y}.$$
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
We almost always talk of a top. space $X$ when we really mean a pair $(X,T_X)$ where $T_X$ is a certain kind of collection of subsets of $X$, called a topology on $X$, and commonly called the collection of open sets. $T_X$ is not determined by $X$ alone, unless $X$ has at most one member. Suppose $T_X$ is a topology on $X$ and $T_Y$ is a topology on $Y,$ and that $Xsupset Csubset Y.$ There may be many possible topologies on the set $C$. Suppose $T_C$ is a topology on $C$ such that $(C,T_C)$ is a sub-space of $(X,T_X)$ and also of $(Y,T_Y)$. By the definition of a sub-space topology, this means that $${xcap C:xin T_X}=T_C={ycap C: yin T_Y}.$$
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
answered Dec 22 '18 at 4:17
DanielWainfleetDanielWainfleet
35.3k31648
35.3k31648
add a comment |
add a comment |
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By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition
$endgroup$
– Ben W
Dec 21 '18 at 15:32
2
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What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions.
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– Henno Brandsma
Dec 21 '18 at 15:33
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I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done.
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– YuiTo Cheng
Dec 21 '18 at 15:33
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@HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..."
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– Steven Wagter
Dec 21 '18 at 15:36
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So the one is not necessarily a subspace of the other.
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– Steven Wagter
Dec 21 '18 at 15:37