Topology closed sets in products of spaces
I got stuck at question 3.7.1 in Bert Mendelson's introduction to Topology.
Prove that a subset $F$ of $X = prod_{i=1}^nX_i$ is closed if and only if F is an intersection of sets, each of which is a finite union of sets of the form $F_1 times F_2 times ... times F_n$, where each $F_i$ is a closed subset of $X_i$. Formulate the corresponding statement in an arbitrary product of topological spaces.
I started with 'F is an intersection ... closed subset of $X_i$' and tried to prove that F must then be closed. Now by this presupposition we can write $F$ as $F = bigcap_{alpha in I} bigcup_{i=1}^m C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i } $. Since we need to prove that the complement of F is open, I tried to take the complement of F.
This way I nearly arrive at the first desired implication; using DeMorgan's rules I obtain
$F = bigcup_{alpha in I} bigcap_{i=1}^m (C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i })^C $. Now I suspect $C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i }$ is a closed subset of $X$ for each $alpha,i$ from which it follows that $F$ is open by the definition of the topology of a product of spaces. However I do not know how to prove this; this isn't true in general.
Does anyone know how to prove this? Or is there something wrong with my approach? After I understand the first implication I'm hoping to be able to do the converse myself.
Please let me know if parts of my question or notation are unclear. Thanks in advance for any help!
general-topology product-space
asked Nov 23 at 19:19
Steven Wagter
928
add a comment |
I got stuck at question 3.7.1 in Bert Mendelson's introduction to Topology.
Prove that a subset $F$ of $X = prod_{i=1}^nX_i$ is closed if and only if F is an intersection of sets, each of which is a finite union of sets of the form $F_1 times F_2 times ... times F_n$, where each $F_i$ is a closed subset of $X_i$. Formulate the corresponding statement in an arbitrary product of topological spaces.
I started with 'F is an intersection ... closed subset of $X_i$' and tried to prove that F must then be closed. Now by this presupposition we can write $F$ as $F = bigcap_{alpha in I} bigcup_{i=1}^m C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i } $. Since we need to prove that the complement of F is open, I tried to take the complement of F.
This way I nearly arrive at the first desired implication; using DeMorgan's rules I obtain
$F = bigcup_{alpha in I} bigcap_{i=1}^m (C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i })^C $. Now I suspect $C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i }$ is a closed subset of $X$ for each $alpha,i$ from which it follows that $F$ is open by the definition of the topology of a product of spaces. However I do not know how to prove this; this isn't true in general.
Does anyone know how to prove this? Or is there something wrong with my approach? After I understand the first implication I'm hoping to be able to do the converse myself.
Please let me know if parts of my question or notation are unclear. Thanks in advance for any help!
general-topology product-space
asked Nov 23 at 19:19
Steven Wagter
928
Can you write the complement of a basis open subset $prod_{i=1}^N O_i$ as such a finite union?
– Henno Brandsma
Nov 24 at 7:47
It's easier still if you knwo the standard subbase for any product. Just apply de Morgan to that representation of open sets.
– Henno Brandsma
Nov 24 at 7:50
add a comment |
I got stuck at question 3.7.1 in Bert Mendelson's introduction to Topology.
Prove that a subset $F$ of $X = prod_{i=1}^nX_i$ is closed if and only if F is an intersection of sets, each of which is a finite union of sets of the form $F_1 times F_2 times ... times F_n$, where each $F_i$ is a closed subset of $X_i$. Formulate the corresponding statement in an arbitrary product of topological spaces.
I started with 'F is an intersection ... closed subset of $X_i$' and tried to prove that F must then be closed. Now by this presupposition we can write $F$ as $F = bigcap_{alpha in I} bigcup_{i=1}^m C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i } $. Since we need to prove that the complement of F is open, I tried to take the complement of F.
This way I nearly arrive at the first desired implication; using DeMorgan's rules I obtain
$F = bigcup_{alpha in I} bigcap_{i=1}^m (C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i })^C $. Now I suspect $C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i }$ is a closed subset of $X$ for each $alpha,i$ from which it follows that $F$ is open by the definition of the topology of a product of spaces. However I do not know how to prove this; this isn't true in general.
Does anyone know how to prove this? Or is there something wrong with my approach? After I understand the first implication I'm hoping to be able to do the converse myself.
Please let me know if parts of my question or notation are unclear. Thanks in advance for any help!
general-topology product-space
asked Nov 23 at 19:19
Steven Wagter
928
I got stuck at question 3.7.1 in Bert Mendelson's introduction to Topology.
Prove that a subset $F$ of $X = prod_{i=1}^nX_i$ is closed if and only if F is an intersection of sets, each of which is a finite union of sets of the form $F_1 times F_2 times ... times F_n$, where each $F_i$ is a closed subset of $X_i$. Formulate the corresponding statement in an arbitrary product of topological spaces.
I started with 'F is an intersection ... closed subset of $X_i$' and tried to prove that F must then be closed. Now by this presupposition we can write $F$ as $F = bigcap_{alpha in I} bigcup_{i=1}^m C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i } $. Since we need to prove that the complement of F is open, I tried to take the complement of F.
This way I nearly arrive at the first desired implication; using DeMorgan's rules I obtain
$F = bigcup_{alpha in I} bigcap_{i=1}^m (C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i })^C $. Now I suspect $C_{1alpha i } times C_{2alpha i } times ...times C_{nalpha i }$ is a closed subset of $X$ for each $alpha,i$ from which it follows that $F$ is open by the definition of the topology of a product of spaces. However I do not know how to prove this; this isn't true in general.
Does anyone know how to prove this? Or is there something wrong with my approach? After I understand the first implication I'm hoping to be able to do the converse myself.
Please let me know if parts of my question or notation are unclear. Thanks in advance for any help!
general-topology product-space
general-topology product-space
asked Nov 23 at 19:19
Steven Wagter
928
asked Nov 23 at 19:19
Steven Wagter
928
edited Nov 24 at 7:16
asked Nov 23 at 19:19
Steven Wagter
928
asked Nov 23 at 19:19
Steven Wagter
928
asked Nov 23 at 19:19
Steven Wagter
928
928
Can you write the complement of a basis open subset $prod_{i=1}^N O_i$ as such a finite union?
– Henno Brandsma
Nov 24 at 7:47
It's easier still if you knwo the standard subbase for any product. Just apply de Morgan to that representation of open sets.
– Henno Brandsma
Nov 24 at 7:50
add a comment |
Can you write the complement of a basis open subset $prod_{i=1}^N O_i$ as such a finite union?
– Henno Brandsma
Nov 24 at 7:47
It's easier still if you knwo the standard subbase for any product. Just apply de Morgan to that representation of open sets.
– Henno Brandsma
Nov 24 at 7:50
Can you write the complement of a basis open subset $prod_{i=1}^N O_i$ as such a finite union?
– Henno Brandsma
Nov 24 at 7:47
Can you write the complement of a basis open subset $prod_{i=1}^N O_i$ as such a finite union?
– Henno Brandsma
Nov 24 at 7:47
It's easier still if you knwo the standard subbase for any product. Just apply de Morgan to that representation of open sets.
– Henno Brandsma
Nov 24 at 7:50
It's easier still if you knwo the standard subbase for any product. Just apply de Morgan to that representation of open sets.
– Henno Brandsma
Nov 24 at 7:50
add a comment |
1 Answer
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In any product space $prod_{iin I} X_i$ all open sets are unions of (finite intersections of "elementary open sets"), where an "elementary open set" is of the form $p_j^{-1}[O]$ where $j in I$ and $p_i: prod_{iin I} X_i to X_j$ is the projection onto the $j$-th space $X_j$, and $O$ is open in $X_j$. We can see these as $$p_j^{-1}[O] = prod_{iin I} O_i$$ where $O_i =X_i$ for all $i neq j$ and $O_j =O$ sometimes denoted (by abuse of notation ) as $$prod_{ineq j} X_i times O$$
The complement of an elementary open set I'll call an elementary closed set, so this is of a similar form with a closed set at the $j$-th coordinate:
$$p_j^{-1}[O]^{complement} = p_j^{-1}[Xsetminus O] = prod_{iin I} F_i text{ with } F_i = X_i text{ for } i neq j text{ and } F_j=Xsetminus O =\
prod_{i neq j} X_i times F text{ where } F = Xsetminus O text{ is closed}$$
By de Morgan we then have that every closed set is the intersection of (finite unions of "elementary closed sets"), which is the form you are looking for.
answered Nov 24 at 8:23
Henno Brandsma
104k346113
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
add a comment |
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In any product space $prod_{iin I} X_i$ all open sets are unions of (finite intersections of "elementary open sets"), where an "elementary open set" is of the form $p_j^{-1}[O]$ where $j in I$ and $p_i: prod_{iin I} X_i to X_j$ is the projection onto the $j$-th space $X_j$, and $O$ is open in $X_j$. We can see these as $$p_j^{-1}[O] = prod_{iin I} O_i$$ where $O_i =X_i$ for all $i neq j$ and $O_j =O$ sometimes denoted (by abuse of notation ) as $$prod_{ineq j} X_i times O$$
The complement of an elementary open set I'll call an elementary closed set, so this is of a similar form with a closed set at the $j$-th coordinate:
$$p_j^{-1}[O]^{complement} = p_j^{-1}[Xsetminus O] = prod_{iin I} F_i text{ with } F_i = X_i text{ for } i neq j text{ and } F_j=Xsetminus O =\
prod_{i neq j} X_i times F text{ where } F = Xsetminus O text{ is closed}$$
By de Morgan we then have that every closed set is the intersection of (finite unions of "elementary closed sets"), which is the form you are looking for.
answered Nov 24 at 8:23
Henno Brandsma
104k346113
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
add a comment |
In any product space $prod_{iin I} X_i$ all open sets are unions of (finite intersections of "elementary open sets"), where an "elementary open set" is of the form $p_j^{-1}[O]$ where $j in I$ and $p_i: prod_{iin I} X_i to X_j$ is the projection onto the $j$-th space $X_j$, and $O$ is open in $X_j$. We can see these as $$p_j^{-1}[O] = prod_{iin I} O_i$$ where $O_i =X_i$ for all $i neq j$ and $O_j =O$ sometimes denoted (by abuse of notation ) as $$prod_{ineq j} X_i times O$$
The complement of an elementary open set I'll call an elementary closed set, so this is of a similar form with a closed set at the $j$-th coordinate:
$$p_j^{-1}[O]^{complement} = p_j^{-1}[Xsetminus O] = prod_{iin I} F_i text{ with } F_i = X_i text{ for } i neq j text{ and } F_j=Xsetminus O =\
prod_{i neq j} X_i times F text{ where } F = Xsetminus O text{ is closed}$$
By de Morgan we then have that every closed set is the intersection of (finite unions of "elementary closed sets"), which is the form you are looking for.
answered Nov 24 at 8:23
Henno Brandsma
104k346113
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
add a comment |
In any product space $prod_{iin I} X_i$ all open sets are unions of (finite intersections of "elementary open sets"), where an "elementary open set" is of the form $p_j^{-1}[O]$ where $j in I$ and $p_i: prod_{iin I} X_i to X_j$ is the projection onto the $j$-th space $X_j$, and $O$ is open in $X_j$. We can see these as $$p_j^{-1}[O] = prod_{iin I} O_i$$ where $O_i =X_i$ for all $i neq j$ and $O_j =O$ sometimes denoted (by abuse of notation ) as $$prod_{ineq j} X_i times O$$
The complement of an elementary open set I'll call an elementary closed set, so this is of a similar form with a closed set at the $j$-th coordinate:
$$p_j^{-1}[O]^{complement} = p_j^{-1}[Xsetminus O] = prod_{iin I} F_i text{ with } F_i = X_i text{ for } i neq j text{ and } F_j=Xsetminus O =\
prod_{i neq j} X_i times F text{ where } F = Xsetminus O text{ is closed}$$
By de Morgan we then have that every closed set is the intersection of (finite unions of "elementary closed sets"), which is the form you are looking for.
answered Nov 24 at 8:23
Henno Brandsma
104k346113
In any product space $prod_{iin I} X_i$ all open sets are unions of (finite intersections of "elementary open sets"), where an "elementary open set" is of the form $p_j^{-1}[O]$ where $j in I$ and $p_i: prod_{iin I} X_i to X_j$ is the projection onto the $j$-th space $X_j$, and $O$ is open in $X_j$. We can see these as $$p_j^{-1}[O] = prod_{iin I} O_i$$ where $O_i =X_i$ for all $i neq j$ and $O_j =O$ sometimes denoted (by abuse of notation ) as $$prod_{ineq j} X_i times O$$
The complement of an elementary open set I'll call an elementary closed set, so this is of a similar form with a closed set at the $j$-th coordinate:
$$p_j^{-1}[O]^{complement} = p_j^{-1}[Xsetminus O] = prod_{iin I} F_i text{ with } F_i = X_i text{ for } i neq j text{ and } F_j=Xsetminus O =\
prod_{i neq j} X_i times F text{ where } F = Xsetminus O text{ is closed}$$
By de Morgan we then have that every closed set is the intersection of (finite unions of "elementary closed sets"), which is the form you are looking for.
answered Nov 24 at 8:23
Henno Brandsma
104k346113
answered Nov 24 at 8:23
Henno Brandsma
104k346113
answered Nov 24 at 8:23
Henno Brandsma
104k346113
answered Nov 24 at 8:23
Henno Brandsma
104k346113
104k346113
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
add a comment |
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
Thank you very much! I understand the closed -> intersection of (finite unions of "elementary closed sets") implication now. For the converse implication I thought this was the correct approach: $F = bigcap_{alpha in I} bigcup_{i=1}^m bigcap_{j=1}^n p_j^{-1}(C_{alpha j}) implies C(F) = bigcup_{alpha in I} bigcap_{i=1}^m bigcup_{j=1}^n p_j^{-1}(C_{alpha j})^C$, where each $p_j^{-1}(C_{alpha j})^C$ open and hence $C(F)$ open and $F$ closed. Is this correct?
– Steven Wagter
Nov 24 at 13:32
add a comment |
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Can you write the complement of a basis open subset $prod_{i=1}^N O_i$ as such a finite union?
– Henno Brandsma
Nov 24 at 7:47
It's easier still if you knwo the standard subbase for any product. Just apply de Morgan to that representation of open sets.
– Henno Brandsma
Nov 24 at 7:50