Equivalence of $T_1$ axiom definition The 2019 Stack Overflow Developer Survey Results Are InWhy is $T_1$ required for a topological space to be $T_4$?T_1 separation axiom in topologyDefinition of $T_1$ Space in Kolmogorov-Fomin Introductory Real Analysis bookEquivalence of Definitions for $T_1$Confusion in an order topology being Hausdorff and $T_1$.Why is the $T_1$ axiom necessary in this lemma?Do sets in a topological space satisfying the $T_1$ axiom need to have infinitely many points?Is this correct definition of $T_1$ space?Prove equivalence of singletons being closed with the $T_1$ property.Equivalent formulation of $T_1$ condition.
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Equivalence of $T_1$ axiom definition
The 2019 Stack Overflow Developer Survey Results Are InWhy is $T_1$ required for a topological space to be $T_4$?T_1 separation axiom in topologyDefinition of $T_1$ Space in Kolmogorov-Fomin Introductory Real Analysis bookEquivalence of Definitions for $T_1$Confusion in an order topology being Hausdorff and $T_1$.Why is the $T_1$ axiom necessary in this lemma?Do sets in a topological space satisfying the $T_1$ axiom need to have infinitely many points?Is this correct definition of $T_1$ space?Prove equivalence of singletons being closed with the $T_1$ property.Equivalent formulation of $T_1$ condition.
$begingroup$
I already know the following definition for $T_1$ axiom:
$(1)$ Let $X$ a topological space. We say $X$ satisties $T_1$ axiom if all the finite subset of $X$ are closed.
Now, I want to prove that definition is equivalent to the following definition:
$(2)$ Let $X$ a topological space. We say $X$ satisfies $T_1$ axiom if for all distincts $x,yin X$, there are neighborhoods $xin U_x, yin U_y$ in $X$ such that $xnotin U_y$ and $ynotin U_x$.
To prove $(1)Rightarrow(2)$, we know that $A:=x,y$ is closed in $X$ for all distincts $x,yin X$. Suppose that for all neighborhood $U_xni x,U_yni y$ we have $xin U_y$ and $yin U_x$. How can I get a contradiction with the fact that $A^c$ is open?
To prove $(2)Rightarrow (1)$, we can show that $x,y$ is closed in $X$ for all distincts $x,yin X$, and that is enough, but I don't know how to do that.
Can someone give me some hints?
general-topology separation-axioms
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
$endgroup$
add a comment |
$begingroup$
I already know the following definition for $T_1$ axiom:
$(1)$ Let $X$ a topological space. We say $X$ satisties $T_1$ axiom if all the finite subset of $X$ are closed.
Now, I want to prove that definition is equivalent to the following definition:
$(2)$ Let $X$ a topological space. We say $X$ satisfies $T_1$ axiom if for all distincts $x,yin X$, there are neighborhoods $xin U_x, yin U_y$ in $X$ such that $xnotin U_y$ and $ynotin U_x$.
To prove $(1)Rightarrow(2)$, we know that $A:=x,y$ is closed in $X$ for all distincts $x,yin X$. Suppose that for all neighborhood $U_xni x,U_yni y$ we have $xin U_y$ and $yin U_x$. How can I get a contradiction with the fact that $A^c$ is open?
To prove $(2)Rightarrow (1)$, we can show that $x,y$ is closed in $X$ for all distincts $x,yin X$, and that is enough, but I don't know how to do that.
Can someone give me some hints?
general-topology separation-axioms
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
$endgroup$
add a comment |
$begingroup$
I already know the following definition for $T_1$ axiom:
$(1)$ Let $X$ a topological space. We say $X$ satisties $T_1$ axiom if all the finite subset of $X$ are closed.
Now, I want to prove that definition is equivalent to the following definition:
$(2)$ Let $X$ a topological space. We say $X$ satisfies $T_1$ axiom if for all distincts $x,yin X$, there are neighborhoods $xin U_x, yin U_y$ in $X$ such that $xnotin U_y$ and $ynotin U_x$.
To prove $(1)Rightarrow(2)$, we know that $A:=x,y$ is closed in $X$ for all distincts $x,yin X$. Suppose that for all neighborhood $U_xni x,U_yni y$ we have $xin U_y$ and $yin U_x$. How can I get a contradiction with the fact that $A^c$ is open?
To prove $(2)Rightarrow (1)$, we can show that $x,y$ is closed in $X$ for all distincts $x,yin X$, and that is enough, but I don't know how to do that.
Can someone give me some hints?
general-topology separation-axioms
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
$endgroup$
I already know the following definition for $T_1$ axiom:
$(1)$ Let $X$ a topological space. We say $X$ satisties $T_1$ axiom if all the finite subset of $X$ are closed.
Now, I want to prove that definition is equivalent to the following definition:
$(2)$ Let $X$ a topological space. We say $X$ satisfies $T_1$ axiom if for all distincts $x,yin X$, there are neighborhoods $xin U_x, yin U_y$ in $X$ such that $xnotin U_y$ and $ynotin U_x$.
To prove $(1)Rightarrow(2)$, we know that $A:=x,y$ is closed in $X$ for all distincts $x,yin X$. Suppose that for all neighborhood $U_xni x,U_yni y$ we have $xin U_y$ and $yin U_x$. How can I get a contradiction with the fact that $A^c$ is open?
To prove $(2)Rightarrow (1)$, we can show that $x,y$ is closed in $X$ for all distincts $x,yin X$, and that is enough, but I don't know how to do that.
Can someone give me some hints?
general-topology separation-axioms
general-topology separation-axioms
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
asked Mar 30 at 21:29
Mateus RochaMateus Rocha
832117
832117
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
$(1)implies(2)$ Take $x,yin X$, with $xneq y$. Since $x$ is closed, $Xsetminusx$ is open and, since $yneq x$, $yin Xsetminusx$. There you have it: $Xsetminusx$ is a neighborhhod of $y$ no which $x$ doesn't belong.
$(2)implies(1)$ If $xin X$, then, for each $yin Xsetminusx$, lete $A_y$ be an open set such that $xnotin A_y$. Then $bigcup_yin XsetminusxA_y=Xsetminusx$. This proves that $Xsetminusx$ is open. So, $x$ is closed. Since this occurs for every singleton…
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
add a comment |
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$begingroup$
$(1)implies(2)$ Take $x,yin X$, with $xneq y$. Since $x$ is closed, $Xsetminusx$ is open and, since $yneq x$, $yin Xsetminusx$. There you have it: $Xsetminusx$ is a neighborhhod of $y$ no which $x$ doesn't belong.
$(2)implies(1)$ If $xin X$, then, for each $yin Xsetminusx$, lete $A_y$ be an open set such that $xnotin A_y$. Then $bigcup_yin XsetminusxA_y=Xsetminusx$. This proves that $Xsetminusx$ is open. So, $x$ is closed. Since this occurs for every singleton…
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
add a comment |
$begingroup$
$(1)implies(2)$ Take $x,yin X$, with $xneq y$. Since $x$ is closed, $Xsetminusx$ is open and, since $yneq x$, $yin Xsetminusx$. There you have it: $Xsetminusx$ is a neighborhhod of $y$ no which $x$ doesn't belong.
$(2)implies(1)$ If $xin X$, then, for each $yin Xsetminusx$, lete $A_y$ be an open set such that $xnotin A_y$. Then $bigcup_yin XsetminusxA_y=Xsetminusx$. This proves that $Xsetminusx$ is open. So, $x$ is closed. Since this occurs for every singleton…
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
add a comment |
$begingroup$
$(1)implies(2)$ Take $x,yin X$, with $xneq y$. Since $x$ is closed, $Xsetminusx$ is open and, since $yneq x$, $yin Xsetminusx$. There you have it: $Xsetminusx$ is a neighborhhod of $y$ no which $x$ doesn't belong.
$(2)implies(1)$ If $xin X$, then, for each $yin Xsetminusx$, lete $A_y$ be an open set such that $xnotin A_y$. Then $bigcup_yin XsetminusxA_y=Xsetminusx$. This proves that $Xsetminusx$ is open. So, $x$ is closed. Since this occurs for every singleton…
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
$endgroup$
$(1)implies(2)$ Take $x,yin X$, with $xneq y$. Since $x$ is closed, $Xsetminusx$ is open and, since $yneq x$, $yin Xsetminusx$. There you have it: $Xsetminusx$ is a neighborhhod of $y$ no which $x$ doesn't belong.
$(2)implies(1)$ If $xin X$, then, for each $yin Xsetminusx$, lete $A_y$ be an open set such that $xnotin A_y$. Then $bigcup_yin XsetminusxA_y=Xsetminusx$. This proves that $Xsetminusx$ is open. So, $x$ is closed. Since this occurs for every singleton…
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
answered Mar 30 at 21:40
José Carlos SantosJosé Carlos Santos
174k23133242
174k23133242
add a comment |
add a comment |
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