A space is Hausdorff iff…
Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:
Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$
Partial proof:
$(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.
$(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?
general-topology separation-axioms
asked 4 hours ago
Ladooscuro
234
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Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:
Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$
Partial proof:
$(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.
$(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?
general-topology separation-axioms
asked 4 hours ago
Ladooscuro
234
New contributor
Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:
Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$
Partial proof:
$(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.
$(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?
general-topology separation-axioms
asked 4 hours ago
Ladooscuro
234
New contributor
Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Here's an exercise from Marco Manetti's "Topologia" book (ex. 3.59 in the italian version) that i'm stuck on:
Prove that a topological space $X$ is $mathrm{T2}iff{x}=bigcaplimits_{Uin,mathcal{I}(x)}kern{-7pt}overline{U};;;forall xin X$
Partial proof:
$(Rightarrow)$ Let $xin X$, suppose that $yin X$ and $xneq y$. They have disjoint neighborhoods because $X$ is $mathrm{T2}$. So $exists A,Bsubset X$ open and disjoint, such that $xin A$ and $yin B$, and as they're disjoint $Asubset B^{kern{1pt}mathrm{c}}$ and $B^{kern{1pt}mathrm{c}}$ is a closed neighborhood of $x$, choosing such a $B_y$ for each $yneq x$ yields a family of closed neighborhoods of $x$, and $bigcaplimits_{yneq x}kern{-3pt}overline{B^{kern{1pt}mathrm{c}}_y}=bigcaplimits_{yneq x}kern{-3pt}B^{kern{1pt}mathrm{c}}_y={x}$.
$(Leftarrow)$ My initial idea was to prove that if in a space $X$ with the property on the right we assume that given a pair of points $x,yin X$ they have no $Ainmathcal{I}(x),Binmathcal{I}(y)$ open and disjoint we have a contradiction, but all my ideas seemed ineffective, how should i proceed?
general-topology separation-axioms
general-topology separation-axioms
asked 4 hours ago
Ladooscuro
234
New contributor
Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ladooscuro
234
New contributor
Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
asked 4 hours ago
Ladooscuro
234
New contributor
Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
Ladooscuro
234
asked 4 hours ago
Ladooscuro
234
234
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Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Ladooscuro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.
Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.
answered 4 hours ago
José Carlos Santos
149k22117219
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
add a comment |
Starting the proof with “let $x,yin X$” is not the right way.
Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.
Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.
answered 3 hours ago
egreg
177k1484200
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
1
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
add a comment |
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2 Answers
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2 Answers
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Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.
Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.
answered 4 hours ago
José Carlos Santos
149k22117219
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
add a comment |
Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.
Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.
answered 4 hours ago
José Carlos Santos
149k22117219
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
add a comment |
Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.
Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.
answered 4 hours ago
José Carlos Santos
149k22117219
Suppose that $X$ is $T_2$. If $x,yin X$, there are $Uinmathcal{I}(x)$ and $Vinmathcal{I}(Y)$ such that $Ucap V=emptyset$. Let $O$ be an open set such that $yin O$ and $Osubset V$. Then $O^complement$ is a closed set that containes $U$. Therefore, $overline Usubset O^complement$ and therefore $ynotinoverline U$. Sinse this occurs for every $yin X$, $bigcap_{Uinmathcal{I}(x)}overline{U}={x}$.
Now, suppose that $X$ is not $T_2$. Take $x,yin X$ such that, for any $Uinmathcal{I}(x)$ and any $Vinmathcal{I}(y)$, $Ucap Vneqemptyset$. It's not hard to prove that $yinbigcap_{Uinmathcal{I}(x)}overline{U}$.
answered 4 hours ago
José Carlos Santos
149k22117219
answered 4 hours ago
José Carlos Santos
149k22117219
answered 4 hours ago
José Carlos Santos
149k22117219
answered 4 hours ago
José Carlos Santos
149k22117219
149k22117219
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
add a comment |
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Thanks for the suggestion!
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Then the proof follows from lemma 3.21 of the book: "x is in the closure of a subset B iff each neighborhood of x has non-empty intersection with B"
– Ladooscuro
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
Indeed it does.
– José Carlos Santos
3 hours ago
add a comment |
Starting the proof with “let $x,yin X$” is not the right way.
Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.
Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.
answered 3 hours ago
egreg
177k1484200
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
1
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
add a comment |
Starting the proof with “let $x,yin X$” is not the right way.
Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.
Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.
answered 3 hours ago
egreg
177k1484200
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
1
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
add a comment |
Starting the proof with “let $x,yin X$” is not the right way.
Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.
Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.
answered 3 hours ago
egreg
177k1484200
Starting the proof with “let $x,yin X$” is not the right way.
Suppose $X$ is Hausdorff and let $xin X$. Suppose $yin X$ and $yne x$. Then there exist $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$ with $Ucap V=emptyset$; in particular $ynotinbar{U}$. Thus $ynotinbigcap_{Uinmathcal{I}(x)}bar{U}$.
Suppose $X$ is not Hausdorff and let $x,yin X$, with $xne y$, such that $Ucap Vneemptyset$, for every $Uinmathcal{I}(x)$ and $Vinmathcal{I}(y)$. In particular, $yinbar{U}$, for every $Uinmathcal{I}(x)$.
answered 3 hours ago
egreg
177k1484200
answered 3 hours ago
egreg
177k1484200
answered 3 hours ago
egreg
177k1484200
answered 3 hours ago
egreg
177k1484200
177k1484200
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
1
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
add a comment |
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
1
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
Care to explain why that's not the right way? except for not assuming x neq y
– Ladooscuro
3 hours ago
1
1
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
@Ladooscuro Well, that's a very important point. Also you want to show something about any $xin X$. So it's better to start with $x$ and introduce the auxiliary $y$ later.
– egreg
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
Thanks, i'll keep it in mind
– Ladooscuro
2 hours ago
add a comment |
Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.
Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.
Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.
Ladooscuro is a new contributor. Be nice, and check out our Code of Conduct.
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