Countable Basis Proof in Munkres
$begingroup$
Munkres writes the following:
Suppose that $X$ has a countable basis. Then there exists a countable subset of $X$ that is dense in $X$.
The proof:
From each nonempty basis element $B_n$, choose a point $x_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$: Given any point $x in X$, every basis element containing $X$ intersects $D$, so $x$ belongs to $overline{D}$.
I don't understand why the bold statement is true. Why can't an open set containing $x$ have empty intersection with $D$?
general-topology
edited Jan 9 at 12:12
Namaste
1
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
$endgroup$
add a comment |
$begingroup$
Munkres writes the following:
Suppose that $X$ has a countable basis. Then there exists a countable subset of $X$ that is dense in $X$.
The proof:
From each nonempty basis element $B_n$, choose a point $x_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$: Given any point $x in X$, every basis element containing $X$ intersects $D$, so $x$ belongs to $overline{D}$.
I don't understand why the bold statement is true. Why can't an open set containing $x$ have empty intersection with $D$?
general-topology
edited Jan 9 at 12:12
Namaste
1
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
$endgroup$
add a comment |
$begingroup$
Munkres writes the following:
Suppose that $X$ has a countable basis. Then there exists a countable subset of $X$ that is dense in $X$.
The proof:
From each nonempty basis element $B_n$, choose a point $x_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$: Given any point $x in X$, every basis element containing $X$ intersects $D$, so $x$ belongs to $overline{D}$.
I don't understand why the bold statement is true. Why can't an open set containing $x$ have empty intersection with $D$?
general-topology
edited Jan 9 at 12:12
Namaste
1
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
$endgroup$
Munkres writes the following:
Suppose that $X$ has a countable basis. Then there exists a countable subset of $X$ that is dense in $X$.
The proof:
From each nonempty basis element $B_n$, choose a point $x_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$: Given any point $x in X$, every basis element containing $X$ intersects $D$, so $x$ belongs to $overline{D}$.
I don't understand why the bold statement is true. Why can't an open set containing $x$ have empty intersection with $D$?
general-topology
general-topology
edited Jan 9 at 12:12
Namaste
1
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
edited Jan 9 at 12:12
Namaste
1
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
edited Jan 9 at 12:12
Namaste
1
edited Jan 9 at 12:12
Namaste
1
edited Jan 9 at 12:12
Namaste
1
1
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
asked Oct 5 '18 at 14:38
yoshiyoshi
1,261917
1,261917
add a comment |
add a comment |
1 Answer
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$begingroup$
If $xin X$, every basis element $U$ containing $x$ is equal to $B_n$, for some natural $n$. But then $x_nin B_n=U$. Since $x_nin D$, this proves that $Ucap Dneqemptyset$. Since every non-empty open set contains some basis element, this proves that every non-empty open set intersects $D$. In other words, $D$ is dense in $X$.
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
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add a comment |
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1 Answer
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$begingroup$
If $xin X$, every basis element $U$ containing $x$ is equal to $B_n$, for some natural $n$. But then $x_nin B_n=U$. Since $x_nin D$, this proves that $Ucap Dneqemptyset$. Since every non-empty open set contains some basis element, this proves that every non-empty open set intersects $D$. In other words, $D$ is dense in $X$.
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
If $xin X$, every basis element $U$ containing $x$ is equal to $B_n$, for some natural $n$. But then $x_nin B_n=U$. Since $x_nin D$, this proves that $Ucap Dneqemptyset$. Since every non-empty open set contains some basis element, this proves that every non-empty open set intersects $D$. In other words, $D$ is dense in $X$.
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
add a comment |
$begingroup$
If $xin X$, every basis element $U$ containing $x$ is equal to $B_n$, for some natural $n$. But then $x_nin B_n=U$. Since $x_nin D$, this proves that $Ucap Dneqemptyset$. Since every non-empty open set contains some basis element, this proves that every non-empty open set intersects $D$. In other words, $D$ is dense in $X$.
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
If $xin X$, every basis element $U$ containing $x$ is equal to $B_n$, for some natural $n$. But then $x_nin B_n=U$. Since $x_nin D$, this proves that $Ucap Dneqemptyset$. Since every non-empty open set contains some basis element, this proves that every non-empty open set intersects $D$. In other words, $D$ is dense in $X$.
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
answered Oct 5 '18 at 14:42
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
add a comment |
add a comment |
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