Mixing
Showing open set
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x in A$ there is an open set $U$ containing $x$ such that $U subset A$. Show that $A$ is open in $X$.
I was given the following definition: A subset $U$ of $X$ is said to be open in $X$ if for each $x in U$ there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U$.
So I did:
Let $x in A$. By hypothesis there is an open set $U$ containing $x$ such that $U subset A$. So since $U$ is open it follows that there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U subset A$. So $A$ is open in $X$.
Could someone let me know what I did wrong? The solution in my book is the same as the answer provided here:
Let X be Topological Space ; $Asubset X$ Suppose that for each $xin A$ there is open set U contain x such that $Usubset A$.Show that A is open in X
general-topology
asked Jul 20 at 17:46
funmath
9918
 |Â
show 1 more comment
up vote
2
down vote
favorite
Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x in A$ there is an open set $U$ containing $x$ such that $U subset A$. Show that $A$ is open in $X$.
I was given the following definition: A subset $U$ of $X$ is said to be open in $X$ if for each $x in U$ there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U$.
So I did:
Let $x in A$. By hypothesis there is an open set $U$ containing $x$ such that $U subset A$. So since $U$ is open it follows that there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U subset A$. So $A$ is open in $X$.
Could someone let me know what I did wrong? The solution in my book is the same as the answer provided here:
Let X be Topological Space ; $Asubset X$ Suppose that for each $xin A$ there is open set U contain x such that $Usubset A$.Show that A is open in X
general-topology
asked Jul 20 at 17:46
funmath
9918
1
Hm. Your proof looks essentially okay, although it looks like you haven't actually defined your basis $mathcalB$ (to be fair, $mathcalB$ is also never defined in your statement of the definition of an open set). Given your definition of an open set, the easiest correction might be to define $mathcalB = mathcalT$, where $mathcalT$ is the topology on $X$.
– mheldman
Jul 20 at 17:54
@mheldman oh yes sorry. By $mathcalB$ I meant a basis for the topology $X$. I’ll add your correction to my answer then. Would you be willing to compare/contrast my answer to the answer in the link I posted? I’ve been googling this question and all answers are essentially the same as the answer in that link. Why do they reach the conclusion that $A$ is a union of open sets?
– funmath
Jul 20 at 18:09
1
Yeah, I assumed you meant that. I think your answer is okay, but the easiest thing would just be to note that $mathcalT$ is a basis for the topology, and since every $xin A$ is contained in some $Uin mathcalT$ where $Usubseteq A$, $A$ is open. It's fine to pick an arbitrary basis (as long as you prove that a basis always exists), but it's usually better to pick a specific basis.
– mheldman
Jul 20 at 18:12
1
To address your confusion about the other answers, another way to show a set is open is to show that it is a union of open sets (as per the definition of a topology). If you haven't proven that $mathcalT$ is a basis for itself (which is straightforward, but adds a few extra lines), then you can avoid doing that by taking this route. Both methods are valid though.
– mheldman
Jul 20 at 18:17
1
I think for a topological space $X$ it is wrong to "define" a set $U$ to be open if for each $xin U$ there is a basis element $Binmathcal B$ such that $xin Bsubseteq U$. You can only speak of a topological space $X$ if a topology $tau_Xsubseteqwp(X)$ is defined. Then a set $U$ is by definition open if $Uintau_X$. In short: to define a set to be open a basis should not be needed.
– drhab
Jul 20 at 18:38
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x in A$ there is an open set $U$ containing $x$ such that $U subset A$. Show that $A$ is open in $X$.
I was given the following definition: A subset $U$ of $X$ is said to be open in $X$ if for each $x in U$ there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U$.
So I did:
Let $x in A$. By hypothesis there is an open set $U$ containing $x$ such that $U subset A$. So since $U$ is open it follows that there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U subset A$. So $A$ is open in $X$.
Could someone let me know what I did wrong? The solution in my book is the same as the answer provided here:
Let X be Topological Space ; $Asubset X$ Suppose that for each $xin A$ there is open set U contain x such that $Usubset A$.Show that A is open in X
general-topology
asked Jul 20 at 17:46
funmath
9918
Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x in A$ there is an open set $U$ containing $x$ such that $U subset A$. Show that $A$ is open in $X$.
I was given the following definition: A subset $U$ of $X$ is said to be open in $X$ if for each $x in U$ there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U$.
So I did:
Let $x in A$. By hypothesis there is an open set $U$ containing $x$ such that $U subset A$. So since $U$ is open it follows that there is a basis element $B in mathcalB$ such that $x in B$ and $B subset U subset A$. So $A$ is open in $X$.
Could someone let me know what I did wrong? The solution in my book is the same as the answer provided here:
Let X be Topological Space ; $Asubset X$ Suppose that for each $xin A$ there is open set U contain x such that $Usubset A$.Show that A is open in X
general-topology
asked Jul 20 at 17:46
funmath
9918
asked Jul 20 at 17:46
funmath
9918
asked Jul 20 at 17:46
funmath
9918
asked Jul 20 at 17:46
funmath
9918
9918
1
Hm. Your proof looks essentially okay, although it looks like you haven't actually defined your basis $mathcalB$ (to be fair, $mathcalB$ is also never defined in your statement of the definition of an open set). Given your definition of an open set, the easiest correction might be to define $mathcalB = mathcalT$, where $mathcalT$ is the topology on $X$.
– mheldman
Jul 20 at 17:54
@mheldman oh yes sorry. By $mathcalB$ I meant a basis for the topology $X$. I’ll add your correction to my answer then. Would you be willing to compare/contrast my answer to the answer in the link I posted? I’ve been googling this question and all answers are essentially the same as the answer in that link. Why do they reach the conclusion that $A$ is a union of open sets?
– funmath
Jul 20 at 18:09
1
Yeah, I assumed you meant that. I think your answer is okay, but the easiest thing would just be to note that $mathcalT$ is a basis for the topology, and since every $xin A$ is contained in some $Uin mathcalT$ where $Usubseteq A$, $A$ is open. It's fine to pick an arbitrary basis (as long as you prove that a basis always exists), but it's usually better to pick a specific basis.
– mheldman
Jul 20 at 18:12
1
To address your confusion about the other answers, another way to show a set is open is to show that it is a union of open sets (as per the definition of a topology). If you haven't proven that $mathcalT$ is a basis for itself (which is straightforward, but adds a few extra lines), then you can avoid doing that by taking this route. Both methods are valid though.
– mheldman
Jul 20 at 18:17
1
I think for a topological space $X$ it is wrong to "define" a set $U$ to be open if for each $xin U$ there is a basis element $Binmathcal B$ such that $xin Bsubseteq U$. You can only speak of a topological space $X$ if a topology $tau_Xsubseteqwp(X)$ is defined. Then a set $U$ is by definition open if $Uintau_X$. In short: to define a set to be open a basis should not be needed.
– drhab
Jul 20 at 18:38
 |Â
show 1 more comment
1
Hm. Your proof looks essentially okay, although it looks like you haven't actually defined your basis $mathcalB$ (to be fair, $mathcalB$ is also never defined in your statement of the definition of an open set). Given your definition of an open set, the easiest correction might be to define $mathcalB = mathcalT$, where $mathcalT$ is the topology on $X$.
– mheldman
Jul 20 at 17:54
@mheldman oh yes sorry. By $mathcalB$ I meant a basis for the topology $X$. I’ll add your correction to my answer then. Would you be willing to compare/contrast my answer to the answer in the link I posted? I’ve been googling this question and all answers are essentially the same as the answer in that link. Why do they reach the conclusion that $A$ is a union of open sets?
– funmath
Jul 20 at 18:09
1
Yeah, I assumed you meant that. I think your answer is okay, but the easiest thing would just be to note that $mathcalT$ is a basis for the topology, and since every $xin A$ is contained in some $Uin mathcalT$ where $Usubseteq A$, $A$ is open. It's fine to pick an arbitrary basis (as long as you prove that a basis always exists), but it's usually better to pick a specific basis.
– mheldman
Jul 20 at 18:12
1
To address your confusion about the other answers, another way to show a set is open is to show that it is a union of open sets (as per the definition of a topology). If you haven't proven that $mathcalT$ is a basis for itself (which is straightforward, but adds a few extra lines), then you can avoid doing that by taking this route. Both methods are valid though.
– mheldman
Jul 20 at 18:17
1
I think for a topological space $X$ it is wrong to "define" a set $U$ to be open if for each $xin U$ there is a basis element $Binmathcal B$ such that $xin Bsubseteq U$. You can only speak of a topological space $X$ if a topology $tau_Xsubseteqwp(X)$ is defined. Then a set $U$ is by definition open if $Uintau_X$. In short: to define a set to be open a basis should not be needed.
– drhab
Jul 20 at 18:38
1
1
Hm. Your proof looks essentially okay, although it looks like you haven't actually defined your basis $mathcalB$ (to be fair, $mathcalB$ is also never defined in your statement of the definition of an open set). Given your definition of an open set, the easiest correction might be to define $mathcalB = mathcalT$, where $mathcalT$ is the topology on $X$.
– mheldman
Jul 20 at 17:54
Hm. Your proof looks essentially okay, although it looks like you haven't actually defined your basis $mathcalB$ (to be fair, $mathcalB$ is also never defined in your statement of the definition of an open set). Given your definition of an open set, the easiest correction might be to define $mathcalB = mathcalT$, where $mathcalT$ is the topology on $X$.
– mheldman
Jul 20 at 17:54
@mheldman oh yes sorry. By $mathcalB$ I meant a basis for the topology $X$. I’ll add your correction to my answer then. Would you be willing to compare/contrast my answer to the answer in the link I posted? I’ve been googling this question and all answers are essentially the same as the answer in that link. Why do they reach the conclusion that $A$ is a union of open sets?
– funmath
Jul 20 at 18:09
@mheldman oh yes sorry. By $mathcalB$ I meant a basis for the topology $X$. I’ll add your correction to my answer then. Would you be willing to compare/contrast my answer to the answer in the link I posted? I’ve been googling this question and all answers are essentially the same as the answer in that link. Why do they reach the conclusion that $A$ is a union of open sets?
– funmath
Jul 20 at 18:09
1
1
Yeah, I assumed you meant that. I think your answer is okay, but the easiest thing would just be to note that $mathcalT$ is a basis for the topology, and since every $xin A$ is contained in some $Uin mathcalT$ where $Usubseteq A$, $A$ is open. It's fine to pick an arbitrary basis (as long as you prove that a basis always exists), but it's usually better to pick a specific basis.
– mheldman
Jul 20 at 18:12
Yeah, I assumed you meant that. I think your answer is okay, but the easiest thing would just be to note that $mathcalT$ is a basis for the topology, and since every $xin A$ is contained in some $Uin mathcalT$ where $Usubseteq A$, $A$ is open. It's fine to pick an arbitrary basis (as long as you prove that a basis always exists), but it's usually better to pick a specific basis.
– mheldman
Jul 20 at 18:12
1
1
To address your confusion about the other answers, another way to show a set is open is to show that it is a union of open sets (as per the definition of a topology). If you haven't proven that $mathcalT$ is a basis for itself (which is straightforward, but adds a few extra lines), then you can avoid doing that by taking this route. Both methods are valid though.
– mheldman
Jul 20 at 18:17
To address your confusion about the other answers, another way to show a set is open is to show that it is a union of open sets (as per the definition of a topology). If you haven't proven that $mathcalT$ is a basis for itself (which is straightforward, but adds a few extra lines), then you can avoid doing that by taking this route. Both methods are valid though.
– mheldman
Jul 20 at 18:17
1
1
I think for a topological space $X$ it is wrong to "define" a set $U$ to be open if for each $xin U$ there is a basis element $Binmathcal B$ such that $xin Bsubseteq U$. You can only speak of a topological space $X$ if a topology $tau_Xsubseteqwp(X)$ is defined. Then a set $U$ is by definition open if $Uintau_X$. In short: to define a set to be open a basis should not be needed.
– drhab
Jul 20 at 18:38
I think for a topological space $X$ it is wrong to "define" a set $U$ to be open if for each $xin U$ there is a basis element $Binmathcal B$ such that $xin Bsubseteq U$. You can only speak of a topological space $X$ if a topology $tau_Xsubseteqwp(X)$ is defined. Then a set $U$ is by definition open if $Uintau_X$. In short: to define a set to be open a basis should not be needed.
– drhab
Jul 20 at 18:38
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
I think you are given a set $X$ and a collection of subsets $mathcalB$ of $X$ satisfying certain properties.
You can then define a topology $mathcalT(mathcalB)$ on $X$ by
$O subseteq X$ is open (i.e. $O in mathcalT(mathcalB)$) iff $forall x in O: exists B_x in mathcalB: x in B_x subseteq O$
which can be shown to be a topology (i.e. obeying the 3 axioms) because $mathcalB$ was not arbitrary but obeyed the special properties.
Now we want to show the simple fact that
$O in mathcalT(mathcalB)$ iff $forall x in O: exists O_x in mathcalT(mathcalB): x in O_x subseteq O$.
The left to right implication is very trivial: for every $x$ we pick $O_x = O$.
The right to left implication is also easy: the right hand side condition implies that $O = bigcup O_x: x in O$ (all $O_x$ are subsets of $O$, hence so is their union, which shows $supseteq$; and every $x$ is in its own $O_x$ which shows $subseteq$) and so $O$ is a union of elements of $mathcalT(mathcalB)$ and so in $mathcalT(mathcalB)$ itself, as topologies are closed under unions.
Note that this proof only needs that $mathcalT(mathcalB)$ is a topology, not how it was defined from the collection $mathcalB$. I assume that this has been checked prior to this.
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think you are given a set $X$ and a collection of subsets $mathcalB$ of $X$ satisfying certain properties.
You can then define a topology $mathcalT(mathcalB)$ on $X$ by
$O subseteq X$ is open (i.e. $O in mathcalT(mathcalB)$) iff $forall x in O: exists B_x in mathcalB: x in B_x subseteq O$
which can be shown to be a topology (i.e. obeying the 3 axioms) because $mathcalB$ was not arbitrary but obeyed the special properties.
Now we want to show the simple fact that
$O in mathcalT(mathcalB)$ iff $forall x in O: exists O_x in mathcalT(mathcalB): x in O_x subseteq O$.
The left to right implication is very trivial: for every $x$ we pick $O_x = O$.
The right to left implication is also easy: the right hand side condition implies that $O = bigcup O_x: x in O$ (all $O_x$ are subsets of $O$, hence so is their union, which shows $supseteq$; and every $x$ is in its own $O_x$ which shows $subseteq$) and so $O$ is a union of elements of $mathcalT(mathcalB)$ and so in $mathcalT(mathcalB)$ itself, as topologies are closed under unions.
Note that this proof only needs that $mathcalT(mathcalB)$ is a topology, not how it was defined from the collection $mathcalB$. I assume that this has been checked prior to this.
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
add a comment |Â
up vote
0
down vote
I think you are given a set $X$ and a collection of subsets $mathcalB$ of $X$ satisfying certain properties.
You can then define a topology $mathcalT(mathcalB)$ on $X$ by
$O subseteq X$ is open (i.e. $O in mathcalT(mathcalB)$) iff $forall x in O: exists B_x in mathcalB: x in B_x subseteq O$
which can be shown to be a topology (i.e. obeying the 3 axioms) because $mathcalB$ was not arbitrary but obeyed the special properties.
Now we want to show the simple fact that
$O in mathcalT(mathcalB)$ iff $forall x in O: exists O_x in mathcalT(mathcalB): x in O_x subseteq O$.
The left to right implication is very trivial: for every $x$ we pick $O_x = O$.
The right to left implication is also easy: the right hand side condition implies that $O = bigcup O_x: x in O$ (all $O_x$ are subsets of $O$, hence so is their union, which shows $supseteq$; and every $x$ is in its own $O_x$ which shows $subseteq$) and so $O$ is a union of elements of $mathcalT(mathcalB)$ and so in $mathcalT(mathcalB)$ itself, as topologies are closed under unions.
Note that this proof only needs that $mathcalT(mathcalB)$ is a topology, not how it was defined from the collection $mathcalB$. I assume that this has been checked prior to this.
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you are given a set $X$ and a collection of subsets $mathcalB$ of $X$ satisfying certain properties.
You can then define a topology $mathcalT(mathcalB)$ on $X$ by
$O subseteq X$ is open (i.e. $O in mathcalT(mathcalB)$) iff $forall x in O: exists B_x in mathcalB: x in B_x subseteq O$
which can be shown to be a topology (i.e. obeying the 3 axioms) because $mathcalB$ was not arbitrary but obeyed the special properties.
Now we want to show the simple fact that
$O in mathcalT(mathcalB)$ iff $forall x in O: exists O_x in mathcalT(mathcalB): x in O_x subseteq O$.
The left to right implication is very trivial: for every $x$ we pick $O_x = O$.
The right to left implication is also easy: the right hand side condition implies that $O = bigcup O_x: x in O$ (all $O_x$ are subsets of $O$, hence so is their union, which shows $supseteq$; and every $x$ is in its own $O_x$ which shows $subseteq$) and so $O$ is a union of elements of $mathcalT(mathcalB)$ and so in $mathcalT(mathcalB)$ itself, as topologies are closed under unions.
Note that this proof only needs that $mathcalT(mathcalB)$ is a topology, not how it was defined from the collection $mathcalB$. I assume that this has been checked prior to this.
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
I think you are given a set $X$ and a collection of subsets $mathcalB$ of $X$ satisfying certain properties.
You can then define a topology $mathcalT(mathcalB)$ on $X$ by
$O subseteq X$ is open (i.e. $O in mathcalT(mathcalB)$) iff $forall x in O: exists B_x in mathcalB: x in B_x subseteq O$
which can be shown to be a topology (i.e. obeying the 3 axioms) because $mathcalB$ was not arbitrary but obeyed the special properties.
Now we want to show the simple fact that
$O in mathcalT(mathcalB)$ iff $forall x in O: exists O_x in mathcalT(mathcalB): x in O_x subseteq O$.
The left to right implication is very trivial: for every $x$ we pick $O_x = O$.
The right to left implication is also easy: the right hand side condition implies that $O = bigcup O_x: x in O$ (all $O_x$ are subsets of $O$, hence so is their union, which shows $supseteq$; and every $x$ is in its own $O_x$ which shows $subseteq$) and so $O$ is a union of elements of $mathcalT(mathcalB)$ and so in $mathcalT(mathcalB)$ itself, as topologies are closed under unions.
Note that this proof only needs that $mathcalT(mathcalB)$ is a topology, not how it was defined from the collection $mathcalB$. I assume that this has been checked prior to this.
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
answered Jul 21 at 5:35
Henno Brandsma
91.6k342100
91.6k342100
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857877%2fshowing-open-set%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Hm. Your proof looks essentially okay, although it looks like you haven't actually defined your basis $mathcalB$ (to be fair, $mathcalB$ is also never defined in your statement of the definition of an open set). Given your definition of an open set, the easiest correction might be to define $mathcalB = mathcalT$, where $mathcalT$ is the topology on $X$.
– mheldman
Jul 20 at 17:54
@mheldman oh yes sorry. By $mathcalB$ I meant a basis for the topology $X$. I’ll add your correction to my answer then. Would you be willing to compare/contrast my answer to the answer in the link I posted? I’ve been googling this question and all answers are essentially the same as the answer in that link. Why do they reach the conclusion that $A$ is a union of open sets?
– funmath
Jul 20 at 18:09
1
Yeah, I assumed you meant that. I think your answer is okay, but the easiest thing would just be to note that $mathcalT$ is a basis for the topology, and since every $xin A$ is contained in some $Uin mathcalT$ where $Usubseteq A$, $A$ is open. It's fine to pick an arbitrary basis (as long as you prove that a basis always exists), but it's usually better to pick a specific basis.
– mheldman
Jul 20 at 18:12
1
To address your confusion about the other answers, another way to show a set is open is to show that it is a union of open sets (as per the definition of a topology). If you haven't proven that $mathcalT$ is a basis for itself (which is straightforward, but adds a few extra lines), then you can avoid doing that by taking this route. Both methods are valid though.
– mheldman
Jul 20 at 18:17
1
I think for a topological space $X$ it is wrong to "define" a set $U$ to be open if for each $xin U$ there is a basis element $Binmathcal B$ such that $xin Bsubseteq U$. You can only speak of a topological space $X$ if a topology $tau_Xsubseteqwp(X)$ is defined. Then a set $U$ is by definition open if $Uintau_X$. In short: to define a set to be open a basis should not be needed.
– drhab
Jul 20 at 18:38