Union of closures
$begingroup$
Let $X$ be a topological space $mathscr{
B}$ be a collection of subsets of $X$. Show that $overline{ bigcup limits_{alpha in mathscr{B}} B_alpha} subset bigcup limits_{alpha in mathscr{B}} overline{B_alpha}$.
My attempt:
If {${B_alpha: alpha in mathscr{B}}$} is a collection of sets in $X$. Then $x in overline{ bigcup limits_{alpha in mathscr{B}} A_alpha}$ then every neighborhood of $U$ of $x$ intersects $bigcup limits_{alpha in mathscr{B}}$ Thus, $U$ must intersects some $B_alpha$, so $x$ must belong to the closure $overline{B_alpha}$ of some $B_alpha$. Therefore, $x in bigcup limits_{alpha in mathscr{B}} overline{B_alpha} $.
general-topology proof-verification
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space $mathscr{
B}$ be a collection of subsets of $X$. Show that $overline{ bigcup limits_{alpha in mathscr{B}} B_alpha} subset bigcup limits_{alpha in mathscr{B}} overline{B_alpha}$.
My attempt:
If {${B_alpha: alpha in mathscr{B}}$} is a collection of sets in $X$. Then $x in overline{ bigcup limits_{alpha in mathscr{B}} A_alpha}$ then every neighborhood of $U$ of $x$ intersects $bigcup limits_{alpha in mathscr{B}}$ Thus, $U$ must intersects some $B_alpha$, so $x$ must belong to the closure $overline{B_alpha}$ of some $B_alpha$. Therefore, $x in bigcup limits_{alpha in mathscr{B}} overline{B_alpha} $.
general-topology proof-verification
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
$endgroup$
add a comment |
$begingroup$
Let $X$ be a topological space $mathscr{
B}$ be a collection of subsets of $X$. Show that $overline{ bigcup limits_{alpha in mathscr{B}} B_alpha} subset bigcup limits_{alpha in mathscr{B}} overline{B_alpha}$.
My attempt:
If {${B_alpha: alpha in mathscr{B}}$} is a collection of sets in $X$. Then $x in overline{ bigcup limits_{alpha in mathscr{B}} A_alpha}$ then every neighborhood of $U$ of $x$ intersects $bigcup limits_{alpha in mathscr{B}}$ Thus, $U$ must intersects some $B_alpha$, so $x$ must belong to the closure $overline{B_alpha}$ of some $B_alpha$. Therefore, $x in bigcup limits_{alpha in mathscr{B}} overline{B_alpha} $.
general-topology proof-verification
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
$endgroup$
Let $X$ be a topological space $mathscr{
B}$ be a collection of subsets of $X$. Show that $overline{ bigcup limits_{alpha in mathscr{B}} B_alpha} subset bigcup limits_{alpha in mathscr{B}} overline{B_alpha}$.
My attempt:
If {${B_alpha: alpha in mathscr{B}}$} is a collection of sets in $X$. Then $x in overline{ bigcup limits_{alpha in mathscr{B}} A_alpha}$ then every neighborhood of $U$ of $x$ intersects $bigcup limits_{alpha in mathscr{B}}$ Thus, $U$ must intersects some $B_alpha$, so $x$ must belong to the closure $overline{B_alpha}$ of some $B_alpha$. Therefore, $x in bigcup limits_{alpha in mathscr{B}} overline{B_alpha} $.
general-topology proof-verification
general-topology proof-verification
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
edited Dec 7 '18 at 13:26
José Carlos Santos
155k22124227
155k22124227
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
asked Dec 7 '18 at 13:22
jaz laykjaz layk
41
41
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your statement is false. Let $Y$ be any non-closed subset of $X$ and assume that $X$ is $T_1$. Then each singleton is closed and $Y=bigcup_{yin Y}{y}=bigcup_{yin Y}overline{{y}}$. However, since $Y$ is not closed, then $overline Ynotsubset Y$. In other words, $overline{bigcup_{yin Y}{y}}notsubsetbigcup_{yin Y}overline{{y}}$.
I suggest that you try to see where is the error in your proof.
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
In my opinion $mathscr{B}$ has to be finite. Without it, the statement is false. Let me give a counterexample:
$$B_k:=[-1,-frac{1}{k}]subsetmathbb{R}$$
Then $cup_k overline{B_k}=cup_k B_k = [-1,0[$, since the $B_k$ are closed. Hence $0$ is not in this union. But
$$0inoverline{cup_k B_k}=[-1,0]$$
If $mathscr{B}$ is finite, then it is correct, since the union of finitely many closed sets is again closed.
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
$endgroup$
add a comment |
$begingroup$
I think there is some mistake in your question.
for example, Take $X=Bbb{R}$ and ${r_n:n in Bbb{N}}$ is an enumeration of $Bbb{Q}$. Now take, $forall n, B_n:={r_n}$.
Clearly, $cl(B_n)={r_n}$ and $cup B_n=Bbb{Q} $, whose closure is $Bbb{R}$. But $cup overline{B_n}=Bbb{Q}$
So, $overline{cup B_n}=Bbb{R}$ whereas $cupoverline{B_n}=Bbb{Q}$
Edit. But the reverse inclusion is true.(Prove it!)
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your statement is false. Let $Y$ be any non-closed subset of $X$ and assume that $X$ is $T_1$. Then each singleton is closed and $Y=bigcup_{yin Y}{y}=bigcup_{yin Y}overline{{y}}$. However, since $Y$ is not closed, then $overline Ynotsubset Y$. In other words, $overline{bigcup_{yin Y}{y}}notsubsetbigcup_{yin Y}overline{{y}}$.
I suggest that you try to see where is the error in your proof.
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
Your statement is false. Let $Y$ be any non-closed subset of $X$ and assume that $X$ is $T_1$. Then each singleton is closed and $Y=bigcup_{yin Y}{y}=bigcup_{yin Y}overline{{y}}$. However, since $Y$ is not closed, then $overline Ynotsubset Y$. In other words, $overline{bigcup_{yin Y}{y}}notsubsetbigcup_{yin Y}overline{{y}}$.
I suggest that you try to see where is the error in your proof.
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
Your statement is false. Let $Y$ be any non-closed subset of $X$ and assume that $X$ is $T_1$. Then each singleton is closed and $Y=bigcup_{yin Y}{y}=bigcup_{yin Y}overline{{y}}$. However, since $Y$ is not closed, then $overline Ynotsubset Y$. In other words, $overline{bigcup_{yin Y}{y}}notsubsetbigcup_{yin Y}overline{{y}}$.
I suggest that you try to see where is the error in your proof.
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
Your statement is false. Let $Y$ be any non-closed subset of $X$ and assume that $X$ is $T_1$. Then each singleton is closed and $Y=bigcup_{yin Y}{y}=bigcup_{yin Y}overline{{y}}$. However, since $Y$ is not closed, then $overline Ynotsubset Y$. In other words, $overline{bigcup_{yin Y}{y}}notsubsetbigcup_{yin Y}overline{{y}}$.
I suggest that you try to see where is the error in your proof.
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
edited Dec 7 '18 at 13:35
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 7 '18 at 13:26
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
add a comment |
add a comment |
$begingroup$
In my opinion $mathscr{B}$ has to be finite. Without it, the statement is false. Let me give a counterexample:
$$B_k:=[-1,-frac{1}{k}]subsetmathbb{R}$$
Then $cup_k overline{B_k}=cup_k B_k = [-1,0[$, since the $B_k$ are closed. Hence $0$ is not in this union. But
$$0inoverline{cup_k B_k}=[-1,0]$$
If $mathscr{B}$ is finite, then it is correct, since the union of finitely many closed sets is again closed.
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
$endgroup$
add a comment |
$begingroup$
In my opinion $mathscr{B}$ has to be finite. Without it, the statement is false. Let me give a counterexample:
$$B_k:=[-1,-frac{1}{k}]subsetmathbb{R}$$
Then $cup_k overline{B_k}=cup_k B_k = [-1,0[$, since the $B_k$ are closed. Hence $0$ is not in this union. But
$$0inoverline{cup_k B_k}=[-1,0]$$
If $mathscr{B}$ is finite, then it is correct, since the union of finitely many closed sets is again closed.
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
$endgroup$
add a comment |
$begingroup$
In my opinion $mathscr{B}$ has to be finite. Without it, the statement is false. Let me give a counterexample:
$$B_k:=[-1,-frac{1}{k}]subsetmathbb{R}$$
Then $cup_k overline{B_k}=cup_k B_k = [-1,0[$, since the $B_k$ are closed. Hence $0$ is not in this union. But
$$0inoverline{cup_k B_k}=[-1,0]$$
If $mathscr{B}$ is finite, then it is correct, since the union of finitely many closed sets is again closed.
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
$endgroup$
In my opinion $mathscr{B}$ has to be finite. Without it, the statement is false. Let me give a counterexample:
$$B_k:=[-1,-frac{1}{k}]subsetmathbb{R}$$
Then $cup_k overline{B_k}=cup_k B_k = [-1,0[$, since the $B_k$ are closed. Hence $0$ is not in this union. But
$$0inoverline{cup_k B_k}=[-1,0]$$
If $mathscr{B}$ is finite, then it is correct, since the union of finitely many closed sets is again closed.
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
answered Dec 7 '18 at 13:35
humanStampedisthumanStampedist
2,203213
2,203213
add a comment |
add a comment |
$begingroup$
I think there is some mistake in your question.
for example, Take $X=Bbb{R}$ and ${r_n:n in Bbb{N}}$ is an enumeration of $Bbb{Q}$. Now take, $forall n, B_n:={r_n}$.
Clearly, $cl(B_n)={r_n}$ and $cup B_n=Bbb{Q} $, whose closure is $Bbb{R}$. But $cup overline{B_n}=Bbb{Q}$
So, $overline{cup B_n}=Bbb{R}$ whereas $cupoverline{B_n}=Bbb{Q}$
Edit. But the reverse inclusion is true.(Prove it!)
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
$endgroup$
add a comment |
$begingroup$
I think there is some mistake in your question.
for example, Take $X=Bbb{R}$ and ${r_n:n in Bbb{N}}$ is an enumeration of $Bbb{Q}$. Now take, $forall n, B_n:={r_n}$.
Clearly, $cl(B_n)={r_n}$ and $cup B_n=Bbb{Q} $, whose closure is $Bbb{R}$. But $cup overline{B_n}=Bbb{Q}$
So, $overline{cup B_n}=Bbb{R}$ whereas $cupoverline{B_n}=Bbb{Q}$
Edit. But the reverse inclusion is true.(Prove it!)
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
$endgroup$
add a comment |
$begingroup$
I think there is some mistake in your question.
for example, Take $X=Bbb{R}$ and ${r_n:n in Bbb{N}}$ is an enumeration of $Bbb{Q}$. Now take, $forall n, B_n:={r_n}$.
Clearly, $cl(B_n)={r_n}$ and $cup B_n=Bbb{Q} $, whose closure is $Bbb{R}$. But $cup overline{B_n}=Bbb{Q}$
So, $overline{cup B_n}=Bbb{R}$ whereas $cupoverline{B_n}=Bbb{Q}$
Edit. But the reverse inclusion is true.(Prove it!)
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
$endgroup$
I think there is some mistake in your question.
for example, Take $X=Bbb{R}$ and ${r_n:n in Bbb{N}}$ is an enumeration of $Bbb{Q}$. Now take, $forall n, B_n:={r_n}$.
Clearly, $cl(B_n)={r_n}$ and $cup B_n=Bbb{Q} $, whose closure is $Bbb{R}$. But $cup overline{B_n}=Bbb{Q}$
So, $overline{cup B_n}=Bbb{R}$ whereas $cupoverline{B_n}=Bbb{Q}$
Edit. But the reverse inclusion is true.(Prove it!)
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
answered Dec 7 '18 at 13:37
Indrajit GhoshIndrajit Ghosh
1,0691718
1,0691718
add a comment |
add a comment |
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