If $f^-1[cap_iin iB_i]subseteq f^-1[B_i]Rightarrow f^-1[cap_iin iB_i]subseteq cap_iin If^-1[B_i]$
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I may be missing something trivial, I manage to prove that
for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$
Now, why can I take the intersection and say that:
$$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$
elementary-set-theory
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up vote
0
down vote
favorite
I may be missing something trivial, I manage to prove that
for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$
Now, why can I take the intersection and say that:
$$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$
elementary-set-theory
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I may be missing something trivial, I manage to prove that
for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$
Now, why can I take the intersection and say that:
$$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$
elementary-set-theory
I may be missing something trivial, I manage to prove that
for all $iin I$ we get $$f^-1[cap_i in IB_i]subseteq f^-1[B_i]$$
Now, why can I take the intersection and say that:
$$f^-1[cap_i in IB_i]subseteq cap_iin If^-1[B_i]$$
elementary-set-theory
edited Aug 15 at 12:23
pointguard0
1,240821
1,240821
asked Aug 15 at 11:49
newhere
792310
792310
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1 Answer
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If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.
Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
1
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.
Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
1
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
add a comment |Â
up vote
2
down vote
accepted
If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.
Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
1
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.
Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$
If you hase a set $S$ and a family of sets $,lambdainLambda$, then, by definition of intersection, asserting that $Ssubsetbigcap_lambdainLambdaS_lambda$ is the same thing as assertiong that $(foralllambdainLambda):Ssubset S_lambda$.
Therefore, since,$$(forall iin I):f^-1left(bigcap_iin IB_iright)subset f^-1(B_i),$$it follows that$$f^-1left(bigcap_iin IB_iright)subsetbigcap_iin If^-1(B_i).$$
answered Aug 15 at 11:55
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José Carlos Santos
116k1699178
116k1699178
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
1
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
add a comment |Â
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
1
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
You clear and precise explation did the trick, thanks!
â newhere
Aug 15 at 15:31
1
1
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
@newhere I'm glad I could help.
â José Carlos Santos
Aug 15 at 16:37
add a comment |Â
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