Mixing
A subset of a metric space is open iff it is a union of open neighborhoods.
$begingroup$
May someone please verify if this proof is correct?
Let E be the union of open neighborhoods, since the union of a collection of open sets is open, it follows that E is an open set. Assume E is an open set therefore, $forall$ p$in$ E $exists$ $r>0$ : $N_r(p)$ $subset$ E. So for $p_1$ , $exists$ $N_{r_1}(p_1)$ $subset$ E and the same thing holds for every p. This means that every element in E has a neighborhood around it, therefore, E is contained within the union of the neighborhoods.
Furthermore, may someone please tell me of possible ways to improve this proof?
real-analysis general-topology metric-spaces
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
$endgroup$
|
show 1 more comment
$begingroup$
May someone please verify if this proof is correct?
Let E be the union of open neighborhoods, since the union of a collection of open sets is open, it follows that E is an open set. Assume E is an open set therefore, $forall$ p$in$ E $exists$ $r>0$ : $N_r(p)$ $subset$ E. So for $p_1$ , $exists$ $N_{r_1}(p_1)$ $subset$ E and the same thing holds for every p. This means that every element in E has a neighborhood around it, therefore, E is contained within the union of the neighborhoods.
Furthermore, may someone please tell me of possible ways to improve this proof?
real-analysis general-topology metric-spaces
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
$endgroup$
$begingroup$
No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls.
$endgroup$
– Lee Mosher
Jan 16 at 2:17
$begingroup$
But is what I said, equivalent to the definition?
$endgroup$
– mathsssislife
Jan 16 at 2:19
$begingroup$
Furthermore, is the union of neighborhoods necessarily a neighborhood?
$endgroup$
– mathsssislife
Jan 16 at 2:20
$begingroup$
The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do.
$endgroup$
– Lee Mosher
Jan 16 at 2:26
$begingroup$
@LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition.
$endgroup$
– mathworker21
Jan 16 at 3:35
|
show 1 more comment
$begingroup$
May someone please verify if this proof is correct?
Let E be the union of open neighborhoods, since the union of a collection of open sets is open, it follows that E is an open set. Assume E is an open set therefore, $forall$ p$in$ E $exists$ $r>0$ : $N_r(p)$ $subset$ E. So for $p_1$ , $exists$ $N_{r_1}(p_1)$ $subset$ E and the same thing holds for every p. This means that every element in E has a neighborhood around it, therefore, E is contained within the union of the neighborhoods.
Furthermore, may someone please tell me of possible ways to improve this proof?
real-analysis general-topology metric-spaces
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
$endgroup$
May someone please verify if this proof is correct?
Let E be the union of open neighborhoods, since the union of a collection of open sets is open, it follows that E is an open set. Assume E is an open set therefore, $forall$ p$in$ E $exists$ $r>0$ : $N_r(p)$ $subset$ E. So for $p_1$ , $exists$ $N_{r_1}(p_1)$ $subset$ E and the same thing holds for every p. This means that every element in E has a neighborhood around it, therefore, E is contained within the union of the neighborhoods.
Furthermore, may someone please tell me of possible ways to improve this proof?
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
asked Jan 16 at 2:13
mathsssislifemathsssislife
408
408
$begingroup$
No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls.
$endgroup$
– Lee Mosher
Jan 16 at 2:17
$begingroup$
But is what I said, equivalent to the definition?
$endgroup$
– mathsssislife
Jan 16 at 2:19
$begingroup$
Furthermore, is the union of neighborhoods necessarily a neighborhood?
$endgroup$
– mathsssislife
Jan 16 at 2:20
$begingroup$
The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do.
$endgroup$
– Lee Mosher
Jan 16 at 2:26
$begingroup$
@LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition.
$endgroup$
– mathworker21
Jan 16 at 3:35
|
show 1 more comment
$begingroup$
No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls.
$endgroup$
– Lee Mosher
Jan 16 at 2:17
$begingroup$
But is what I said, equivalent to the definition?
$endgroup$
– mathsssislife
Jan 16 at 2:19
$begingroup$
Furthermore, is the union of neighborhoods necessarily a neighborhood?
$endgroup$
– mathsssislife
Jan 16 at 2:20
$begingroup$
The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do.
$endgroup$
– Lee Mosher
Jan 16 at 2:26
$begingroup$
@LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition.
$endgroup$
– mathworker21
Jan 16 at 3:35
$begingroup$
No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls.
$endgroup$
– Lee Mosher
Jan 16 at 2:17
$begingroup$
No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls.
$endgroup$
– Lee Mosher
Jan 16 at 2:17
$begingroup$
But is what I said, equivalent to the definition?
$endgroup$
– mathsssislife
Jan 16 at 2:19
$begingroup$
But is what I said, equivalent to the definition?
$endgroup$
– mathsssislife
Jan 16 at 2:19
$begingroup$
Furthermore, is the union of neighborhoods necessarily a neighborhood?
$endgroup$
– mathsssislife
Jan 16 at 2:20
$begingroup$
Furthermore, is the union of neighborhoods necessarily a neighborhood?
$endgroup$
– mathsssislife
Jan 16 at 2:20
$begingroup$
The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do.
$endgroup$
– Lee Mosher
Jan 16 at 2:26
$begingroup$
The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do.
$endgroup$
– Lee Mosher
Jan 16 at 2:26
$begingroup$
@LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition.
$endgroup$
– mathworker21
Jan 16 at 3:35
$begingroup$
@LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition.
$endgroup$
– mathworker21
Jan 16 at 3:35
|
show 1 more comment
1 Answer
1
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$begingroup$
You asked if it is possible to improve your proof that an open set is the union of open sets.
Notation. Let $(M,rho)$ be a metric space. Fix $x_0in M$. Fix $r>0$.
"$B(x_0,r)$" is notation for "${xin M:rho(x,x_0)<r}$."
Terminology. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. Say $S$ is open if for each $xin S$, there exists $r>0$ such that $B(x,r)subseteq S$.
Proposition. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. The following are equivalent.
- (i) $S$ is open.
- (ii) $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets.
Proof.
($Rightarrow$) Assume $S$ is open. Then $S=bigcup_{xin S}B(x,r_x)$, where $r_x>0$ and $B(x,r_x)subseteq S$ for every $xin S$. Because $B(x,r_x)$ is open for every $xin S$, we are done.
($Leftarrow$) Assume $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets. To prove $S$ is open, fix $x_0in S$. Then $x_0in U_{alpha_0}$ for some $alpha_0in A$. Because $U_{alpha_0}$ is open, there exists $r_0>0$ such that $B(x_0,r_0)subseteq U_{alpha_0}$. Because $S=bigcup_{alphain A}U_alpha$, it follows that $B(x_0,r_0)subseteq S$. As a result, $S$ is open.
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
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add a comment |
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1 Answer
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$begingroup$
You asked if it is possible to improve your proof that an open set is the union of open sets.
Notation. Let $(M,rho)$ be a metric space. Fix $x_0in M$. Fix $r>0$.
"$B(x_0,r)$" is notation for "${xin M:rho(x,x_0)<r}$."
Terminology. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. Say $S$ is open if for each $xin S$, there exists $r>0$ such that $B(x,r)subseteq S$.
Proposition. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. The following are equivalent.
- (i) $S$ is open.
- (ii) $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets.
Proof.
($Rightarrow$) Assume $S$ is open. Then $S=bigcup_{xin S}B(x,r_x)$, where $r_x>0$ and $B(x,r_x)subseteq S$ for every $xin S$. Because $B(x,r_x)$ is open for every $xin S$, we are done.
($Leftarrow$) Assume $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets. To prove $S$ is open, fix $x_0in S$. Then $x_0in U_{alpha_0}$ for some $alpha_0in A$. Because $U_{alpha_0}$ is open, there exists $r_0>0$ such that $B(x_0,r_0)subseteq U_{alpha_0}$. Because $S=bigcup_{alphain A}U_alpha$, it follows that $B(x_0,r_0)subseteq S$. As a result, $S$ is open.
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
$endgroup$
add a comment |
$begingroup$
You asked if it is possible to improve your proof that an open set is the union of open sets.
Notation. Let $(M,rho)$ be a metric space. Fix $x_0in M$. Fix $r>0$.
"$B(x_0,r)$" is notation for "${xin M:rho(x,x_0)<r}$."
Terminology. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. Say $S$ is open if for each $xin S$, there exists $r>0$ such that $B(x,r)subseteq S$.
Proposition. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. The following are equivalent.
- (i) $S$ is open.
- (ii) $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets.
Proof.
($Rightarrow$) Assume $S$ is open. Then $S=bigcup_{xin S}B(x,r_x)$, where $r_x>0$ and $B(x,r_x)subseteq S$ for every $xin S$. Because $B(x,r_x)$ is open for every $xin S$, we are done.
($Leftarrow$) Assume $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets. To prove $S$ is open, fix $x_0in S$. Then $x_0in U_{alpha_0}$ for some $alpha_0in A$. Because $U_{alpha_0}$ is open, there exists $r_0>0$ such that $B(x_0,r_0)subseteq U_{alpha_0}$. Because $S=bigcup_{alphain A}U_alpha$, it follows that $B(x_0,r_0)subseteq S$. As a result, $S$ is open.
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
$endgroup$
add a comment |
$begingroup$
You asked if it is possible to improve your proof that an open set is the union of open sets.
Notation. Let $(M,rho)$ be a metric space. Fix $x_0in M$. Fix $r>0$.
"$B(x_0,r)$" is notation for "${xin M:rho(x,x_0)<r}$."
Terminology. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. Say $S$ is open if for each $xin S$, there exists $r>0$ such that $B(x,r)subseteq S$.
Proposition. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. The following are equivalent.
- (i) $S$ is open.
- (ii) $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets.
Proof.
($Rightarrow$) Assume $S$ is open. Then $S=bigcup_{xin S}B(x,r_x)$, where $r_x>0$ and $B(x,r_x)subseteq S$ for every $xin S$. Because $B(x,r_x)$ is open for every $xin S$, we are done.
($Leftarrow$) Assume $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets. To prove $S$ is open, fix $x_0in S$. Then $x_0in U_{alpha_0}$ for some $alpha_0in A$. Because $U_{alpha_0}$ is open, there exists $r_0>0$ such that $B(x_0,r_0)subseteq U_{alpha_0}$. Because $S=bigcup_{alphain A}U_alpha$, it follows that $B(x_0,r_0)subseteq S$. As a result, $S$ is open.
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
$endgroup$
You asked if it is possible to improve your proof that an open set is the union of open sets.
Notation. Let $(M,rho)$ be a metric space. Fix $x_0in M$. Fix $r>0$.
"$B(x_0,r)$" is notation for "${xin M:rho(x,x_0)<r}$."
Terminology. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. Say $S$ is open if for each $xin S$, there exists $r>0$ such that $B(x,r)subseteq S$.
Proposition. Let $(M,rho)$ be a metric space. Fix $Ssubseteq M$. The following are equivalent.
- (i) $S$ is open.
- (ii) $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets.
Proof.
($Rightarrow$) Assume $S$ is open. Then $S=bigcup_{xin S}B(x,r_x)$, where $r_x>0$ and $B(x,r_x)subseteq S$ for every $xin S$. Because $B(x,r_x)$ is open for every $xin S$, we are done.
($Leftarrow$) Assume $S=bigcup_{alphain A}U_alpha$, where ${U_alpha}_{alphain A}$ is a family of open sets. To prove $S$ is open, fix $x_0in S$. Then $x_0in U_{alpha_0}$ for some $alpha_0in A$. Because $U_{alpha_0}$ is open, there exists $r_0>0$ such that $B(x_0,r_0)subseteq U_{alpha_0}$. Because $S=bigcup_{alphain A}U_alpha$, it follows that $B(x_0,r_0)subseteq S$. As a result, $S$ is open.
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
answered Jan 16 at 3:38
Alberto TakaseAlberto Takase
2,260619
2,260619
add a comment |
add a comment |
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$begingroup$
No proof is needed, because the very definition of an open set in a metric space is a set which is a union of open balls.
$endgroup$
– Lee Mosher
Jan 16 at 2:17
$begingroup$
But is what I said, equivalent to the definition?
$endgroup$
– mathsssislife
Jan 16 at 2:19
$begingroup$
Furthermore, is the union of neighborhoods necessarily a neighborhood?
$endgroup$
– mathsssislife
Jan 16 at 2:20
$begingroup$
The definition of an open set, similar to any mathematical definition, allows you to automatically translate back and forth between the two statements "$E$ is open" and "$E$ is a union of open balls", without any proof. That's what definitions do.
$endgroup$
– Lee Mosher
Jan 16 at 2:26
$begingroup$
@LeeMosher that's not the definition Rudin gives. And it's pretty clear from the question that mathsssislife isn't using your definition.
$endgroup$
– mathworker21
Jan 16 at 3:35