A space is Hausdorff iff…
$begingroup$
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 Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
$endgroup$
add a comment |
$begingroup$
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 Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
$endgroup$
add a comment |
$begingroup$
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 Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
$endgroup$
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 Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
asked Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
edited Dec 27 '18 at 21:39
Ladooscuro
asked Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
asked Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
asked Dec 27 '18 at 20:12
LadooscuroLadooscuro
284
284
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
add a comment |
$begingroup$
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 Dec 27 '18 at 21:16
egregegreg
184k1486205
$endgroup$
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
1
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
add a comment |
$begingroup$
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 Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
add a comment |
$begingroup$
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 Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
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 Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
answered Dec 27 '18 at 20:22
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
add a comment |
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
Thanks for the suggestion!
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:35
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
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"
$endgroup$
– Ladooscuro
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
$begingroup$
Indeed it does.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 20:45
add a comment |
$begingroup$
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 Dec 27 '18 at 21:16
egregegreg
184k1486205
$endgroup$
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
1
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
add a comment |
$begingroup$
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 Dec 27 '18 at 21:16
egregegreg
184k1486205
$endgroup$
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
1
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
add a comment |
$begingroup$
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 Dec 27 '18 at 21:16
egregegreg
184k1486205
$endgroup$
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 Dec 27 '18 at 21:16
egregegreg
184k1486205
answered Dec 27 '18 at 21:16
egregegreg
184k1486205
answered Dec 27 '18 at 21:16
egregegreg
184k1486205
answered Dec 27 '18 at 21:16
egregegreg
184k1486205
184k1486205
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
1
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
add a comment |
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
1
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
$begingroup$
Care to explain why that's not the right way? except for not assuming x neq y
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:24
1
1
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
@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.
$endgroup$
– egreg
Dec 27 '18 at 21:36
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
$begingroup$
Thanks, i'll keep it in mind
$endgroup$
– Ladooscuro
Dec 27 '18 at 21:37
add a comment |
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