Set Theory: Union over sets satisfying criterion, rigorous definition?
$begingroup$
I'm trying come to a rigorous understanding of certain types of set unions. For purposes of this question I'm comfortable with set unions written as
$$
mathcal{A} = bigcup A = {x:exists_a(xin a wedge ain A)}
$$
but I am not comfortable with unions expressed using other types of notations..
I am trying to understand Munkres Lemma 13.1 concerning a characterization for the basis ($mathcal{B}$) of a topology*. In fact, I am stuck at the same point as the asker in this question: Proof of Lemma 13.1 in Munkres.
We are considering an open set $U$ and the proof has a step which says for each $x$ in $U$ we know there is a set $B_x$ satisfying $xin B_xsubset U$ and $B_x in mathcal{B}$. I understand and agree with this statement. The next statement is the problem. That statement says that thus
$$
U = bigcup_{xin U}B_x
$$
Let
Intuitively I very much understand this statement. We've found a $B_x$ for each $x$ and we combine together all of the $B_x$ that we found. Let
$$
C = bigcup_{xin U}B_x
$$
Clearly each $x$ in $U$ will be in this collection so $Usubset C$and clearly each element of this collection is a subset of $U$ so $Csubset U$ so clearly $C=U$.
My problem is understanding what is the rigorous definition of the set $C$ in terms of the definition for infinitary unions I have given above. That is, is there some set $b$ such that I can write
$$
C = bigcup b
$$
Where $b$ somehow captures the notion of having one set $B_xsubset U$ inside of it for each $x$ in $U$? How would I construct the set $b$ formally? It seems like it is something like constructing a choice function? (something I'm not really familiar with..)
edit:
Just a few notes to explain along what lines I am thinking about the problem. It seems like I want something like
$$
f:U rightarrow mathcal{B}
$$
with $x in f(x) subset U$. Then I would want $b$ to be the range of $f$ which is guaranteed to be a set by the axiom schema of replacement. The problems are 1) I don't know exactly how to construct the function $f$ and 2) I'm not sure if this is an overkill solution to the issue I am describing when there is in fact a much more straightforward definition..
*However my question doesn't really have to do with this particular proof but rather a general notion about set unions so I don't want to get bogged down in the details of this particular proof.
general-topology elementary-set-theory
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying come to a rigorous understanding of certain types of set unions. For purposes of this question I'm comfortable with set unions written as
$$
mathcal{A} = bigcup A = {x:exists_a(xin a wedge ain A)}
$$
but I am not comfortable with unions expressed using other types of notations..
I am trying to understand Munkres Lemma 13.1 concerning a characterization for the basis ($mathcal{B}$) of a topology*. In fact, I am stuck at the same point as the asker in this question: Proof of Lemma 13.1 in Munkres.
We are considering an open set $U$ and the proof has a step which says for each $x$ in $U$ we know there is a set $B_x$ satisfying $xin B_xsubset U$ and $B_x in mathcal{B}$. I understand and agree with this statement. The next statement is the problem. That statement says that thus
$$
U = bigcup_{xin U}B_x
$$
Let
Intuitively I very much understand this statement. We've found a $B_x$ for each $x$ and we combine together all of the $B_x$ that we found. Let
$$
C = bigcup_{xin U}B_x
$$
Clearly each $x$ in $U$ will be in this collection so $Usubset C$and clearly each element of this collection is a subset of $U$ so $Csubset U$ so clearly $C=U$.
My problem is understanding what is the rigorous definition of the set $C$ in terms of the definition for infinitary unions I have given above. That is, is there some set $b$ such that I can write
$$
C = bigcup b
$$
Where $b$ somehow captures the notion of having one set $B_xsubset U$ inside of it for each $x$ in $U$? How would I construct the set $b$ formally? It seems like it is something like constructing a choice function? (something I'm not really familiar with..)
edit:
Just a few notes to explain along what lines I am thinking about the problem. It seems like I want something like
$$
f:U rightarrow mathcal{B}
$$
with $x in f(x) subset U$. Then I would want $b$ to be the range of $f$ which is guaranteed to be a set by the axiom schema of replacement. The problems are 1) I don't know exactly how to construct the function $f$ and 2) I'm not sure if this is an overkill solution to the issue I am describing when there is in fact a much more straightforward definition..
*However my question doesn't really have to do with this particular proof but rather a general notion about set unions so I don't want to get bogged down in the details of this particular proof.
general-topology elementary-set-theory
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
$endgroup$
$begingroup$
I don't know much topology, but it seems like unless you have a formula to pick an element from the set ${ain{cal B}mid xin aland asubseteq U}$ for arbitrary $x$, you will have to use choice function
$endgroup$
– Holo
Jan 2 at 7:44
$begingroup$
@Holo: That is correct. In many cases, however, doing the greedy thing and just taking all suitable sets works just as well. (Not always, though.)
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:45
$begingroup$
So is it correct to say that the notation used in Munkres assumes the axiom of choice to complete the proof?
$endgroup$
– Jagerber48
Jan 2 at 7:49
$begingroup$
@AsafKaragila it looks like in this case the OP just need minimal cover(the union to be minimal, not the family) from the basis, so it looks like it is suitable method
$endgroup$
– Holo
Jan 2 at 7:52
$begingroup$
It is safe to assume that any sane person learning topology for the first time should not be concerned with the axiom of choice and just use it.
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:53
|
show 1 more comment
$begingroup$
I'm trying come to a rigorous understanding of certain types of set unions. For purposes of this question I'm comfortable with set unions written as
$$
mathcal{A} = bigcup A = {x:exists_a(xin a wedge ain A)}
$$
but I am not comfortable with unions expressed using other types of notations..
I am trying to understand Munkres Lemma 13.1 concerning a characterization for the basis ($mathcal{B}$) of a topology*. In fact, I am stuck at the same point as the asker in this question: Proof of Lemma 13.1 in Munkres.
We are considering an open set $U$ and the proof has a step which says for each $x$ in $U$ we know there is a set $B_x$ satisfying $xin B_xsubset U$ and $B_x in mathcal{B}$. I understand and agree with this statement. The next statement is the problem. That statement says that thus
$$
U = bigcup_{xin U}B_x
$$
Let
Intuitively I very much understand this statement. We've found a $B_x$ for each $x$ and we combine together all of the $B_x$ that we found. Let
$$
C = bigcup_{xin U}B_x
$$
Clearly each $x$ in $U$ will be in this collection so $Usubset C$and clearly each element of this collection is a subset of $U$ so $Csubset U$ so clearly $C=U$.
My problem is understanding what is the rigorous definition of the set $C$ in terms of the definition for infinitary unions I have given above. That is, is there some set $b$ such that I can write
$$
C = bigcup b
$$
Where $b$ somehow captures the notion of having one set $B_xsubset U$ inside of it for each $x$ in $U$? How would I construct the set $b$ formally? It seems like it is something like constructing a choice function? (something I'm not really familiar with..)
edit:
Just a few notes to explain along what lines I am thinking about the problem. It seems like I want something like
$$
f:U rightarrow mathcal{B}
$$
with $x in f(x) subset U$. Then I would want $b$ to be the range of $f$ which is guaranteed to be a set by the axiom schema of replacement. The problems are 1) I don't know exactly how to construct the function $f$ and 2) I'm not sure if this is an overkill solution to the issue I am describing when there is in fact a much more straightforward definition..
*However my question doesn't really have to do with this particular proof but rather a general notion about set unions so I don't want to get bogged down in the details of this particular proof.
general-topology elementary-set-theory
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
$endgroup$
I'm trying come to a rigorous understanding of certain types of set unions. For purposes of this question I'm comfortable with set unions written as
$$
mathcal{A} = bigcup A = {x:exists_a(xin a wedge ain A)}
$$
but I am not comfortable with unions expressed using other types of notations..
I am trying to understand Munkres Lemma 13.1 concerning a characterization for the basis ($mathcal{B}$) of a topology*. In fact, I am stuck at the same point as the asker in this question: Proof of Lemma 13.1 in Munkres.
We are considering an open set $U$ and the proof has a step which says for each $x$ in $U$ we know there is a set $B_x$ satisfying $xin B_xsubset U$ and $B_x in mathcal{B}$. I understand and agree with this statement. The next statement is the problem. That statement says that thus
$$
U = bigcup_{xin U}B_x
$$
Let
Intuitively I very much understand this statement. We've found a $B_x$ for each $x$ and we combine together all of the $B_x$ that we found. Let
$$
C = bigcup_{xin U}B_x
$$
Clearly each $x$ in $U$ will be in this collection so $Usubset C$and clearly each element of this collection is a subset of $U$ so $Csubset U$ so clearly $C=U$.
My problem is understanding what is the rigorous definition of the set $C$ in terms of the definition for infinitary unions I have given above. That is, is there some set $b$ such that I can write
$$
C = bigcup b
$$
Where $b$ somehow captures the notion of having one set $B_xsubset U$ inside of it for each $x$ in $U$? How would I construct the set $b$ formally? It seems like it is something like constructing a choice function? (something I'm not really familiar with..)
edit:
Just a few notes to explain along what lines I am thinking about the problem. It seems like I want something like
$$
f:U rightarrow mathcal{B}
$$
with $x in f(x) subset U$. Then I would want $b$ to be the range of $f$ which is guaranteed to be a set by the axiom schema of replacement. The problems are 1) I don't know exactly how to construct the function $f$ and 2) I'm not sure if this is an overkill solution to the issue I am describing when there is in fact a much more straightforward definition..
*However my question doesn't really have to do with this particular proof but rather a general notion about set unions so I don't want to get bogged down in the details of this particular proof.
general-topology elementary-set-theory
general-topology elementary-set-theory
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
edited Jan 2 at 9:11
Andrés E. Caicedo
65.8k8160251
65.8k8160251
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
asked Jan 2 at 7:14
Jagerber48Jagerber48
14810
14810
$begingroup$
I don't know much topology, but it seems like unless you have a formula to pick an element from the set ${ain{cal B}mid xin aland asubseteq U}$ for arbitrary $x$, you will have to use choice function
$endgroup$
– Holo
Jan 2 at 7:44
$begingroup$
@Holo: That is correct. In many cases, however, doing the greedy thing and just taking all suitable sets works just as well. (Not always, though.)
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:45
$begingroup$
So is it correct to say that the notation used in Munkres assumes the axiom of choice to complete the proof?
$endgroup$
– Jagerber48
Jan 2 at 7:49
$begingroup$
@AsafKaragila it looks like in this case the OP just need minimal cover(the union to be minimal, not the family) from the basis, so it looks like it is suitable method
$endgroup$
– Holo
Jan 2 at 7:52
$begingroup$
It is safe to assume that any sane person learning topology for the first time should not be concerned with the axiom of choice and just use it.
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:53
|
show 1 more comment
$begingroup$
I don't know much topology, but it seems like unless you have a formula to pick an element from the set ${ain{cal B}mid xin aland asubseteq U}$ for arbitrary $x$, you will have to use choice function
$endgroup$
– Holo
Jan 2 at 7:44
$begingroup$
@Holo: That is correct. In many cases, however, doing the greedy thing and just taking all suitable sets works just as well. (Not always, though.)
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:45
$begingroup$
So is it correct to say that the notation used in Munkres assumes the axiom of choice to complete the proof?
$endgroup$
– Jagerber48
Jan 2 at 7:49
$begingroup$
@AsafKaragila it looks like in this case the OP just need minimal cover(the union to be minimal, not the family) from the basis, so it looks like it is suitable method
$endgroup$
– Holo
Jan 2 at 7:52
$begingroup$
It is safe to assume that any sane person learning topology for the first time should not be concerned with the axiom of choice and just use it.
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:53
$begingroup$
I don't know much topology, but it seems like unless you have a formula to pick an element from the set ${ain{cal B}mid xin aland asubseteq U}$ for arbitrary $x$, you will have to use choice function
$endgroup$
– Holo
Jan 2 at 7:44
$begingroup$
I don't know much topology, but it seems like unless you have a formula to pick an element from the set ${ain{cal B}mid xin aland asubseteq U}$ for arbitrary $x$, you will have to use choice function
$endgroup$
– Holo
Jan 2 at 7:44
$begingroup$
@Holo: That is correct. In many cases, however, doing the greedy thing and just taking all suitable sets works just as well. (Not always, though.)
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:45
$begingroup$
@Holo: That is correct. In many cases, however, doing the greedy thing and just taking all suitable sets works just as well. (Not always, though.)
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:45
$begingroup$
So is it correct to say that the notation used in Munkres assumes the axiom of choice to complete the proof?
$endgroup$
– Jagerber48
Jan 2 at 7:49
$begingroup$
So is it correct to say that the notation used in Munkres assumes the axiom of choice to complete the proof?
$endgroup$
– Jagerber48
Jan 2 at 7:49
$begingroup$
@AsafKaragila it looks like in this case the OP just need minimal cover(the union to be minimal, not the family) from the basis, so it looks like it is suitable method
$endgroup$
– Holo
Jan 2 at 7:52
$begingroup$
@AsafKaragila it looks like in this case the OP just need minimal cover(the union to be minimal, not the family) from the basis, so it looks like it is suitable method
$endgroup$
– Holo
Jan 2 at 7:52
$begingroup$
It is safe to assume that any sane person learning topology for the first time should not be concerned with the axiom of choice and just use it.
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:53
$begingroup$
It is safe to assume that any sane person learning topology for the first time should not be concerned with the axiom of choice and just use it.
$endgroup$
– Asaf Karagila♦
Jan 2 at 7:53
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
You can take
$$
b = {Bin mathcal{B}colon Bsubseteq U}.
$$
Obviously, $$C = bigcup bsubseteq U.$$
For each $xin U$ there exists some $B_xin b$ such that $xin B_x$, hence $xin C$.
You don't have to make any choices. You should just take more sets in $b$ than you really need.
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
$endgroup$
$begingroup$
Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
$endgroup$
– Jagerber48
Jan 2 at 7:35
$begingroup$
Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
$endgroup$
– Jagerber48
Jan 2 at 7:38
$begingroup$
In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
$endgroup$
– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
$endgroup$
– Jagerber48
Jan 2 at 7:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can take
$$
b = {Bin mathcal{B}colon Bsubseteq U}.
$$
Obviously, $$C = bigcup bsubseteq U.$$
For each $xin U$ there exists some $B_xin b$ such that $xin B_x$, hence $xin C$.
You don't have to make any choices. You should just take more sets in $b$ than you really need.
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
$endgroup$
$begingroup$
Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
$endgroup$
– Jagerber48
Jan 2 at 7:35
$begingroup$
Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
$endgroup$
– Jagerber48
Jan 2 at 7:38
$begingroup$
In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
$endgroup$
– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
$endgroup$
– Jagerber48
Jan 2 at 7:44
add a comment |
$begingroup$
You can take
$$
b = {Bin mathcal{B}colon Bsubseteq U}.
$$
Obviously, $$C = bigcup bsubseteq U.$$
For each $xin U$ there exists some $B_xin b$ such that $xin B_x$, hence $xin C$.
You don't have to make any choices. You should just take more sets in $b$ than you really need.
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
$endgroup$
$begingroup$
Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
$endgroup$
– Jagerber48
Jan 2 at 7:35
$begingroup$
Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
$endgroup$
– Jagerber48
Jan 2 at 7:38
$begingroup$
In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
$endgroup$
– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
$endgroup$
– Jagerber48
Jan 2 at 7:44
add a comment |
$begingroup$
You can take
$$
b = {Bin mathcal{B}colon Bsubseteq U}.
$$
Obviously, $$C = bigcup bsubseteq U.$$
For each $xin U$ there exists some $B_xin b$ such that $xin B_x$, hence $xin C$.
You don't have to make any choices. You should just take more sets in $b$ than you really need.
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
$endgroup$
You can take
$$
b = {Bin mathcal{B}colon Bsubseteq U}.
$$
Obviously, $$C = bigcup bsubseteq U.$$
For each $xin U$ there exists some $B_xin b$ such that $xin B_x$, hence $xin C$.
You don't have to make any choices. You should just take more sets in $b$ than you really need.
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
answered Jan 2 at 7:31
Ilya VlasovIlya Vlasov
170213
170213
$begingroup$
Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
$endgroup$
– Jagerber48
Jan 2 at 7:35
$begingroup$
Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
$endgroup$
– Jagerber48
Jan 2 at 7:38
$begingroup$
In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
$endgroup$
– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
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– Jagerber48
Jan 2 at 7:44
add a comment |
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Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
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– Jagerber48
Jan 2 at 7:35
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Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
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– Jagerber48
Jan 2 at 7:38
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In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
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– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
$endgroup$
– Jagerber48
Jan 2 at 7:44
$begingroup$
Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
$endgroup$
– Jagerber48
Jan 2 at 7:35
$begingroup$
Yes I think I noticed this solution. It is important to point out not only that for each $x in U$ there is some $B_xin b$ with $xin B_x$ but also that this $B_xsubset U$ so that $B_x in b$. This is a technicality though. I like this because it answers my question but I'm curious if there is a way to construct a set $b$ which more closely follows the logic of specifically picking one $B_x$ for each $x$ and then collects them into a set $b$.
$endgroup$
– Jagerber48
Jan 2 at 7:35
$begingroup$
Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
$endgroup$
– Jagerber48
Jan 2 at 7:38
$begingroup$
Sorry, the first part of that last comment didn't quite make sense about $B_x subset U$. In any case I still agree this set $b$ would work for what I am trying to do but I'm still curious if there is one closer to the spirit of the above proof as I asked for in my previous comment.
$endgroup$
– Jagerber48
Jan 2 at 7:38
$begingroup$
In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
$endgroup$
– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
In most cases "picking" involves usage of some version of choice. It's not good to use it, when you can avoid it. If you assume the axiom of choice, you can construct $b$ as in the answer above.
$endgroup$
– Ilya Vlasov
Jan 2 at 7:38
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
$endgroup$
– Jagerber48
Jan 2 at 7:44
$begingroup$
Yes, this was a little bit my concern and why I was asking. I wondered if the Munkres proof implicitly assume some version of choice to work out the way it did.. However, I'm curious if you can construct the set without resorting to some sort of axiom because the existence of the sets $B_x$ that you need are already guaranteed to exist because of the topology axioms (as pointed out by @mathcounterexample.net in their answer.
$endgroup$
– Jagerber48
Jan 2 at 7:44
add a comment |
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$begingroup$
I don't know much topology, but it seems like unless you have a formula to pick an element from the set ${ain{cal B}mid xin aland asubseteq U}$ for arbitrary $x$, you will have to use choice function
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– Holo
Jan 2 at 7:44
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@Holo: That is correct. In many cases, however, doing the greedy thing and just taking all suitable sets works just as well. (Not always, though.)
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– Asaf Karagila♦
Jan 2 at 7:45
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So is it correct to say that the notation used in Munkres assumes the axiom of choice to complete the proof?
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– Jagerber48
Jan 2 at 7:49
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@AsafKaragila it looks like in this case the OP just need minimal cover(the union to be minimal, not the family) from the basis, so it looks like it is suitable method
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– Holo
Jan 2 at 7:52
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It is safe to assume that any sane person learning topology for the first time should not be concerned with the axiom of choice and just use it.
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– Asaf Karagila♦
Jan 2 at 7:53