Can we find $V_n$ in such a way that $bigcaplimits_{n=1}^{infty} V_n$ is countable?
$begingroup$
If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.
My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?
Any help is appreciated. Thank you.
real-analysis baire-category
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
$endgroup$
add a comment |
$begingroup$
If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.
My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?
Any help is appreciated. Thank you.
real-analysis baire-category
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
$endgroup$
add a comment |
$begingroup$
If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.
My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?
Any help is appreciated. Thank you.
real-analysis baire-category
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
$endgroup$
If $V_n$ is open, dense subset of $mathbb R$ for each $nin mathbb N$ then by Baire's Category theorem we know that $displaystyle cap_{n=1}^{infty} V_n$ is dense in $mathbb R$.
My question is that : Can we find $V_n$ in such a way that $displaystyle bigcap_{n=1}^{infty} V_n$ is countable set?
Any help is appreciated. Thank you.
real-analysis baire-category
real-analysis baire-category
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
edited Dec 29 '18 at 9:59
Martin Sleziak
44.7k9117272
44.7k9117272
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
asked Dec 20 '18 at 13:00
nurun neshanurun nesha
1,0222623
1,0222623
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
add a comment |
$begingroup$
No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
add a comment |
$begingroup$
No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
add a comment |
$begingroup$
No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
No, we cannot. Suppose otherwise. Then $bigcap_{ninmathbb N}V_n={r_1,r_2,r_3,ldots}$. Now, let $W_n=V_nsetminus{r_n}$. Then $W_n$ is also an open dense subset of $mathbb R$. But now $bigcap_{ninmathbb N}W_n=emptyset$, which is impossible, by Baire's theorem.
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
edited Dec 20 '18 at 13:21
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 20 '18 at 13:07
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
add a comment |
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
We could also take $W_n=V_n setminus {r_n}$.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:12
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
@nurunnesha You are right. That's even simpler. Do you mind if I edit my answer so that it uses your idea?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 13:13
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
$begingroup$
Yeah, sure edit your answer.
$endgroup$
– nurun nesha
Dec 20 '18 at 13:16
add a comment |
$begingroup$
No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
$begingroup$
No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
$begingroup$
No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
$endgroup$
No. If $S = int_{n=1}^infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
answered Dec 20 '18 at 13:06
Robert IsraelRobert Israel
321k23210463
321k23210463
add a comment |
add a comment |
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