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Let $(X_i, p_i)_i in I$ be a collection of based Hausdorff spaces. Show that $bigvee_i in IX_i$ is Hausdorff
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Let $(X_i, p_i)_i in I$ be a collection of based Hausdorff spaces. Show that $bigvee_i in IX_i$ is Hausdorff
I tried to prove this but ended up getting stuck. Basically the idea I had was the following. By definition we have that $$bigvee_i in IX_i = bigsqcup_i in IX_i / p_i $$ and we know that $bigsqcup_i in IX_i$ is Hausdorff. So now pick two points $x, y in bigvee_i in IX_i$. Then consider the quotient map $$q : bigsqcup_i in IX_i to bigsqcup_i in IX_i / p_i .$$
I was thinking that since all quotient maps are open maps I could choose elements $alpha in q^-1[x]$ and $beta in q^-1[y]$ and then select disjoint open neighborhoods $U$ and $V$ of $alpha$ and $beta$ respectively and then show that $q[U] cap q[V]$ is empty (and that way construct the required neighborhoods to show Hausdorfness). But I ran into a roadblock because I have no way to ensure that $q[U] cap q[V] = emptyset$ . For example if $U$ contains a base point $p_j$ and $V$ contains a base point $p_k$ then I'd have $[p_i] subseteq q[U] cap q[V] $.
How can I go about proving the above?
general-topology algebraic-topology
asked Jul 16 at 14:21
Perturbative
3,53111039
add a comment |Â
up vote
1
down vote
favorite
Let $(X_i, p_i)_i in I$ be a collection of based Hausdorff spaces. Show that $bigvee_i in IX_i$ is Hausdorff
I tried to prove this but ended up getting stuck. Basically the idea I had was the following. By definition we have that $$bigvee_i in IX_i = bigsqcup_i in IX_i / p_i $$ and we know that $bigsqcup_i in IX_i$ is Hausdorff. So now pick two points $x, y in bigvee_i in IX_i$. Then consider the quotient map $$q : bigsqcup_i in IX_i to bigsqcup_i in IX_i / p_i .$$
I was thinking that since all quotient maps are open maps I could choose elements $alpha in q^-1[x]$ and $beta in q^-1[y]$ and then select disjoint open neighborhoods $U$ and $V$ of $alpha$ and $beta$ respectively and then show that $q[U] cap q[V]$ is empty (and that way construct the required neighborhoods to show Hausdorfness). But I ran into a roadblock because I have no way to ensure that $q[U] cap q[V] = emptyset$ . For example if $U$ contains a base point $p_j$ and $V$ contains a base point $p_k$ then I'd have $[p_i] subseteq q[U] cap q[V] $.
How can I go about proving the above?
general-topology algebraic-topology
asked Jul 16 at 14:21
Perturbative
3,53111039
"..since all quotient maps are open maps.." That is not true. The converse is true.
– drhab
Jul 16 at 14:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(X_i, p_i)_i in I$ be a collection of based Hausdorff spaces. Show that $bigvee_i in IX_i$ is Hausdorff
I tried to prove this but ended up getting stuck. Basically the idea I had was the following. By definition we have that $$bigvee_i in IX_i = bigsqcup_i in IX_i / p_i $$ and we know that $bigsqcup_i in IX_i$ is Hausdorff. So now pick two points $x, y in bigvee_i in IX_i$. Then consider the quotient map $$q : bigsqcup_i in IX_i to bigsqcup_i in IX_i / p_i .$$
I was thinking that since all quotient maps are open maps I could choose elements $alpha in q^-1[x]$ and $beta in q^-1[y]$ and then select disjoint open neighborhoods $U$ and $V$ of $alpha$ and $beta$ respectively and then show that $q[U] cap q[V]$ is empty (and that way construct the required neighborhoods to show Hausdorfness). But I ran into a roadblock because I have no way to ensure that $q[U] cap q[V] = emptyset$ . For example if $U$ contains a base point $p_j$ and $V$ contains a base point $p_k$ then I'd have $[p_i] subseteq q[U] cap q[V] $.
How can I go about proving the above?
general-topology algebraic-topology
asked Jul 16 at 14:21
Perturbative
3,53111039
Let $(X_i, p_i)_i in I$ be a collection of based Hausdorff spaces. Show that $bigvee_i in IX_i$ is Hausdorff
I tried to prove this but ended up getting stuck. Basically the idea I had was the following. By definition we have that $$bigvee_i in IX_i = bigsqcup_i in IX_i / p_i $$ and we know that $bigsqcup_i in IX_i$ is Hausdorff. So now pick two points $x, y in bigvee_i in IX_i$. Then consider the quotient map $$q : bigsqcup_i in IX_i to bigsqcup_i in IX_i / p_i .$$
I was thinking that since all quotient maps are open maps I could choose elements $alpha in q^-1[x]$ and $beta in q^-1[y]$ and then select disjoint open neighborhoods $U$ and $V$ of $alpha$ and $beta$ respectively and then show that $q[U] cap q[V]$ is empty (and that way construct the required neighborhoods to show Hausdorfness). But I ran into a roadblock because I have no way to ensure that $q[U] cap q[V] = emptyset$ . For example if $U$ contains a base point $p_j$ and $V$ contains a base point $p_k$ then I'd have $[p_i] subseteq q[U] cap q[V] $.
How can I go about proving the above?
general-topology algebraic-topology
asked Jul 16 at 14:21
Perturbative
3,53111039
asked Jul 16 at 14:21
Perturbative
3,53111039
asked Jul 16 at 14:21
Perturbative
3,53111039
asked Jul 16 at 14:21
Perturbative
3,53111039
3,53111039
"..since all quotient maps are open maps.." That is not true. The converse is true.
– drhab
Jul 16 at 14:32
add a comment |Â
"..since all quotient maps are open maps.." That is not true. The converse is true.
– drhab
Jul 16 at 14:32
"..since all quotient maps are open maps.." That is not true. The converse is true.
– drhab
Jul 16 at 14:32
"..since all quotient maps are open maps.." That is not true. The converse is true.
– drhab
Jul 16 at 14:32
add a comment |Â
1 Answer
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1
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You've got the right idea. Just do it by points. Pick distinct points $a, b$ in your wedge. Suppose neither one is the basepoint. If $a in X_i$ then we can separate $a$ from $p_i$ by assumption. Do the same for $b in X_j$. Now it's clear those those open sets around $a$ and $b$ respectively remain open in the wedge (because they avoid all basepoints by design) and are still disjoint. That case is now done.
Now suppose that $a$ is the basepoint in the wedge and $b$ is not. If $b in X_j$ then separate $p_j$ from $b$ in $X_j$. This is given by, say, open sets $U$ and $V$ in $X_j$ with $p_j in U$ and $b in V$. You're then tempted to say that the images of $U$ and $V$ in the wedge do the trick, but that's not so because $q(U)$ probably isn't open: its inverse image contains $U$ and all the basepoints $p_i$ ($i in I$). Those other basepoints may not be open in their respective $X_i$.
The fix is simple. $q(V)$ is still open in the wedge (and contains the "original" $b$) as it doesn't hit the basepoint, so we're fine there. For the open set about $a$ in the wedge, take
$$
left( bigcup_i neq jq(X_i) right) cup q(U).
$$
This clearly contains $a$ and avoids $q(V)$, but is it open? Yes, because its preimage under the quotient map is $left(bigcup_i neq j X_iright) cup U$ which is open in the disjoint union (before identifying basepoints). The point is that the preimage of $q(U)$ isn't $U$ but is instead $U cup p_i mid i in I$, but all those basepoints get gobbled up in the open $X_i$, so the total preimage is open as claimed.
answered Jul 16 at 15:13
Randall
7,2471825
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You've got the right idea. Just do it by points. Pick distinct points $a, b$ in your wedge. Suppose neither one is the basepoint. If $a in X_i$ then we can separate $a$ from $p_i$ by assumption. Do the same for $b in X_j$. Now it's clear those those open sets around $a$ and $b$ respectively remain open in the wedge (because they avoid all basepoints by design) and are still disjoint. That case is now done.
Now suppose that $a$ is the basepoint in the wedge and $b$ is not. If $b in X_j$ then separate $p_j$ from $b$ in $X_j$. This is given by, say, open sets $U$ and $V$ in $X_j$ with $p_j in U$ and $b in V$. You're then tempted to say that the images of $U$ and $V$ in the wedge do the trick, but that's not so because $q(U)$ probably isn't open: its inverse image contains $U$ and all the basepoints $p_i$ ($i in I$). Those other basepoints may not be open in their respective $X_i$.
The fix is simple. $q(V)$ is still open in the wedge (and contains the "original" $b$) as it doesn't hit the basepoint, so we're fine there. For the open set about $a$ in the wedge, take
$$
left( bigcup_i neq jq(X_i) right) cup q(U).
$$
This clearly contains $a$ and avoids $q(V)$, but is it open? Yes, because its preimage under the quotient map is $left(bigcup_i neq j X_iright) cup U$ which is open in the disjoint union (before identifying basepoints). The point is that the preimage of $q(U)$ isn't $U$ but is instead $U cup p_i mid i in I$, but all those basepoints get gobbled up in the open $X_i$, so the total preimage is open as claimed.
answered Jul 16 at 15:13
Randall
7,2471825
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
add a comment |Â
up vote
1
down vote
accepted
You've got the right idea. Just do it by points. Pick distinct points $a, b$ in your wedge. Suppose neither one is the basepoint. If $a in X_i$ then we can separate $a$ from $p_i$ by assumption. Do the same for $b in X_j$. Now it's clear those those open sets around $a$ and $b$ respectively remain open in the wedge (because they avoid all basepoints by design) and are still disjoint. That case is now done.
Now suppose that $a$ is the basepoint in the wedge and $b$ is not. If $b in X_j$ then separate $p_j$ from $b$ in $X_j$. This is given by, say, open sets $U$ and $V$ in $X_j$ with $p_j in U$ and $b in V$. You're then tempted to say that the images of $U$ and $V$ in the wedge do the trick, but that's not so because $q(U)$ probably isn't open: its inverse image contains $U$ and all the basepoints $p_i$ ($i in I$). Those other basepoints may not be open in their respective $X_i$.
The fix is simple. $q(V)$ is still open in the wedge (and contains the "original" $b$) as it doesn't hit the basepoint, so we're fine there. For the open set about $a$ in the wedge, take
$$
left( bigcup_i neq jq(X_i) right) cup q(U).
$$
This clearly contains $a$ and avoids $q(V)$, but is it open? Yes, because its preimage under the quotient map is $left(bigcup_i neq j X_iright) cup U$ which is open in the disjoint union (before identifying basepoints). The point is that the preimage of $q(U)$ isn't $U$ but is instead $U cup p_i mid i in I$, but all those basepoints get gobbled up in the open $X_i$, so the total preimage is open as claimed.
answered Jul 16 at 15:13
Randall
7,2471825
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You've got the right idea. Just do it by points. Pick distinct points $a, b$ in your wedge. Suppose neither one is the basepoint. If $a in X_i$ then we can separate $a$ from $p_i$ by assumption. Do the same for $b in X_j$. Now it's clear those those open sets around $a$ and $b$ respectively remain open in the wedge (because they avoid all basepoints by design) and are still disjoint. That case is now done.
Now suppose that $a$ is the basepoint in the wedge and $b$ is not. If $b in X_j$ then separate $p_j$ from $b$ in $X_j$. This is given by, say, open sets $U$ and $V$ in $X_j$ with $p_j in U$ and $b in V$. You're then tempted to say that the images of $U$ and $V$ in the wedge do the trick, but that's not so because $q(U)$ probably isn't open: its inverse image contains $U$ and all the basepoints $p_i$ ($i in I$). Those other basepoints may not be open in their respective $X_i$.
The fix is simple. $q(V)$ is still open in the wedge (and contains the "original" $b$) as it doesn't hit the basepoint, so we're fine there. For the open set about $a$ in the wedge, take
$$
left( bigcup_i neq jq(X_i) right) cup q(U).
$$
This clearly contains $a$ and avoids $q(V)$, but is it open? Yes, because its preimage under the quotient map is $left(bigcup_i neq j X_iright) cup U$ which is open in the disjoint union (before identifying basepoints). The point is that the preimage of $q(U)$ isn't $U$ but is instead $U cup p_i mid i in I$, but all those basepoints get gobbled up in the open $X_i$, so the total preimage is open as claimed.
answered Jul 16 at 15:13
Randall
7,2471825
You've got the right idea. Just do it by points. Pick distinct points $a, b$ in your wedge. Suppose neither one is the basepoint. If $a in X_i$ then we can separate $a$ from $p_i$ by assumption. Do the same for $b in X_j$. Now it's clear those those open sets around $a$ and $b$ respectively remain open in the wedge (because they avoid all basepoints by design) and are still disjoint. That case is now done.
Now suppose that $a$ is the basepoint in the wedge and $b$ is not. If $b in X_j$ then separate $p_j$ from $b$ in $X_j$. This is given by, say, open sets $U$ and $V$ in $X_j$ with $p_j in U$ and $b in V$. You're then tempted to say that the images of $U$ and $V$ in the wedge do the trick, but that's not so because $q(U)$ probably isn't open: its inverse image contains $U$ and all the basepoints $p_i$ ($i in I$). Those other basepoints may not be open in their respective $X_i$.
The fix is simple. $q(V)$ is still open in the wedge (and contains the "original" $b$) as it doesn't hit the basepoint, so we're fine there. For the open set about $a$ in the wedge, take
$$
left( bigcup_i neq jq(X_i) right) cup q(U).
$$
This clearly contains $a$ and avoids $q(V)$, but is it open? Yes, because its preimage under the quotient map is $left(bigcup_i neq j X_iright) cup U$ which is open in the disjoint union (before identifying basepoints). The point is that the preimage of $q(U)$ isn't $U$ but is instead $U cup p_i mid i in I$, but all those basepoints get gobbled up in the open $X_i$, so the total preimage is open as claimed.
answered Jul 16 at 15:13
Randall
7,2471825
answered Jul 16 at 15:13
Randall
7,2471825
answered Jul 16 at 15:13
Randall
7,2471825
answered Jul 16 at 15:13
Randall
7,2471825
7,2471825
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
add a comment |Â
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
What do you mean when you say "If $a in X_i$"? Because we're picking $a in bigvee_alpha in A X_alpha$. Perhaps you meant the picking $a$ in the image of $X_i$ (under the quotient map $q$) which is a subset of the wedge. But in that case we can't "seperate $a$ from $p_i$" because to do that would assume Hausdorffness of the image of $X_i$ in the wedge.
– Perturbative
Jul 22 at 16:48
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
Most people don’t make a distinction between the term and its image in their writing. This is standard abuse of language.
– Randall
Jul 22 at 17:35
add a comment |Â
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"..since all quotient maps are open maps.." That is not true. The converse is true.
– drhab
Jul 16 at 14:32