Understanding the closure operation in the topology generated by $(a,infty),a in mathbbR$
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Consider the basis $B = (a,infty),a in mathbbR$ for a topology in $mathbbR$. I've already proven that this is a basis, verifying that its union is the whole $mathbbR$ and for every $x in B_1 cap B_2$, there is $B_3 in B$ such that $x in B_3 subseteq B_1 cap B_2$.
Now I have to describe the interior and closure operations in this topology.
Let $A=(a,infty)$ (Why is it sufficient to analyze for a set $A$ in this form?)
Interior is the union of all open sets contained in $A$, and since $tau = calU in B$ is the topology, then $A^o=A=(a,infty)$. (This looks fine to me, is it correct?)
Closure is the intersection of all closed sets that contain $A$, then $barA = A cup a = (a,infty)cupa$ (I really don't get this. If the closure is a closed set, then its complement must be open, however $barA^c = (-infty,a)$, which doesn't look like can be a union of elements of $B$).
I tried to understand this solution by Michael de Guzman. (Exercise 5A-3)
Please answer my questions above in parenthesis.
Thanks.
https://michaeltdeguzman.files.wordpress.com/2012/02/topohw32.pdf1
general-topology proof-explanation
asked Sep 10 at 20:00
dude3221
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Consider the basis $B = (a,infty),a in mathbbR$ for a topology in $mathbbR$. I've already proven that this is a basis, verifying that its union is the whole $mathbbR$ and for every $x in B_1 cap B_2$, there is $B_3 in B$ such that $x in B_3 subseteq B_1 cap B_2$.
Now I have to describe the interior and closure operations in this topology.
Let $A=(a,infty)$ (Why is it sufficient to analyze for a set $A$ in this form?)
Interior is the union of all open sets contained in $A$, and since $tau = calU in B$ is the topology, then $A^o=A=(a,infty)$. (This looks fine to me, is it correct?)
Closure is the intersection of all closed sets that contain $A$, then $barA = A cup a = (a,infty)cupa$ (I really don't get this. If the closure is a closed set, then its complement must be open, however $barA^c = (-infty,a)$, which doesn't look like can be a union of elements of $B$).
I tried to understand this solution by Michael de Guzman. (Exercise 5A-3)
Please answer my questions above in parenthesis.
Thanks.
https://michaeltdeguzman.files.wordpress.com/2012/02/topohw32.pdf1
general-topology proof-explanation
asked Sep 10 at 20:00
dude3221
39013
1
Per Henno Brandsma's answer below, here's another way to see that $overline(a,infty)$ is the entire real line. Check that $B cup emptyset, mathbbR$ is a topological space. So the only closed sets in the topology are $emptyset$, $mathbbR$, and complements of elements of $B$, The only closed set containing $(a, infty)$, therefore, is $mathbbR$.
– Dabbler
Sep 10 at 21:30
1
My previous comment also answers your other two questions. Since $B cup emptyset, mathbbR$ is a topological space, you can restrict your attention to sets of the form $A = (a, infty)$ when discussing interior and closure operations; the remaining open sets are trivial. The answer to your second question is "yes", if you add cases for $emptyset$ and $mathbbR$. And to your third question, $(infty, a)$ is not open in the topology, and so you don't need to write it as a union of the basis elements.
– Dabbler
Sep 10 at 21:43
add a comment |Â
up vote
0
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favorite
up vote
0
down vote
favorite
Consider the basis $B = (a,infty),a in mathbbR$ for a topology in $mathbbR$. I've already proven that this is a basis, verifying that its union is the whole $mathbbR$ and for every $x in B_1 cap B_2$, there is $B_3 in B$ such that $x in B_3 subseteq B_1 cap B_2$.
Now I have to describe the interior and closure operations in this topology.
Let $A=(a,infty)$ (Why is it sufficient to analyze for a set $A$ in this form?)
Interior is the union of all open sets contained in $A$, and since $tau = calU in B$ is the topology, then $A^o=A=(a,infty)$. (This looks fine to me, is it correct?)
Closure is the intersection of all closed sets that contain $A$, then $barA = A cup a = (a,infty)cupa$ (I really don't get this. If the closure is a closed set, then its complement must be open, however $barA^c = (-infty,a)$, which doesn't look like can be a union of elements of $B$).
I tried to understand this solution by Michael de Guzman. (Exercise 5A-3)
Please answer my questions above in parenthesis.
Thanks.
https://michaeltdeguzman.files.wordpress.com/2012/02/topohw32.pdf1
general-topology proof-explanation
asked Sep 10 at 20:00
dude3221
39013
Consider the basis $B = (a,infty),a in mathbbR$ for a topology in $mathbbR$. I've already proven that this is a basis, verifying that its union is the whole $mathbbR$ and for every $x in B_1 cap B_2$, there is $B_3 in B$ such that $x in B_3 subseteq B_1 cap B_2$.
Now I have to describe the interior and closure operations in this topology.
Let $A=(a,infty)$ (Why is it sufficient to analyze for a set $A$ in this form?)
Interior is the union of all open sets contained in $A$, and since $tau = calU in B$ is the topology, then $A^o=A=(a,infty)$. (This looks fine to me, is it correct?)
Closure is the intersection of all closed sets that contain $A$, then $barA = A cup a = (a,infty)cupa$ (I really don't get this. If the closure is a closed set, then its complement must be open, however $barA^c = (-infty,a)$, which doesn't look like can be a union of elements of $B$).
I tried to understand this solution by Michael de Guzman. (Exercise 5A-3)
Please answer my questions above in parenthesis.
Thanks.
https://michaeltdeguzman.files.wordpress.com/2012/02/topohw32.pdf1
general-topology proof-explanation
general-topology proof-explanation
asked Sep 10 at 20:00
dude3221
39013
asked Sep 10 at 20:00
dude3221
39013
asked Sep 10 at 20:00
dude3221
39013
asked Sep 10 at 20:00
dude3221
39013
asked Sep 10 at 20:00
dude3221
39013
39013
1
Per Henno Brandsma's answer below, here's another way to see that $overline(a,infty)$ is the entire real line. Check that $B cup emptyset, mathbbR$ is a topological space. So the only closed sets in the topology are $emptyset$, $mathbbR$, and complements of elements of $B$, The only closed set containing $(a, infty)$, therefore, is $mathbbR$.
– Dabbler
Sep 10 at 21:30
1
My previous comment also answers your other two questions. Since $B cup emptyset, mathbbR$ is a topological space, you can restrict your attention to sets of the form $A = (a, infty)$ when discussing interior and closure operations; the remaining open sets are trivial. The answer to your second question is "yes", if you add cases for $emptyset$ and $mathbbR$. And to your third question, $(infty, a)$ is not open in the topology, and so you don't need to write it as a union of the basis elements.
– Dabbler
Sep 10 at 21:43
add a comment |Â
1
Per Henno Brandsma's answer below, here's another way to see that $overline(a,infty)$ is the entire real line. Check that $B cup emptyset, mathbbR$ is a topological space. So the only closed sets in the topology are $emptyset$, $mathbbR$, and complements of elements of $B$, The only closed set containing $(a, infty)$, therefore, is $mathbbR$.
– Dabbler
Sep 10 at 21:30
1
My previous comment also answers your other two questions. Since $B cup emptyset, mathbbR$ is a topological space, you can restrict your attention to sets of the form $A = (a, infty)$ when discussing interior and closure operations; the remaining open sets are trivial. The answer to your second question is "yes", if you add cases for $emptyset$ and $mathbbR$. And to your third question, $(infty, a)$ is not open in the topology, and so you don't need to write it as a union of the basis elements.
– Dabbler
Sep 10 at 21:43
1
1
Per Henno Brandsma's answer below, here's another way to see that $overline(a,infty)$ is the entire real line. Check that $B cup emptyset, mathbbR$ is a topological space. So the only closed sets in the topology are $emptyset$, $mathbbR$, and complements of elements of $B$, The only closed set containing $(a, infty)$, therefore, is $mathbbR$.
– Dabbler
Sep 10 at 21:30
Per Henno Brandsma's answer below, here's another way to see that $overline(a,infty)$ is the entire real line. Check that $B cup emptyset, mathbbR$ is a topological space. So the only closed sets in the topology are $emptyset$, $mathbbR$, and complements of elements of $B$, The only closed set containing $(a, infty)$, therefore, is $mathbbR$.
– Dabbler
Sep 10 at 21:30
1
1
My previous comment also answers your other two questions. Since $B cup emptyset, mathbbR$ is a topological space, you can restrict your attention to sets of the form $A = (a, infty)$ when discussing interior and closure operations; the remaining open sets are trivial. The answer to your second question is "yes", if you add cases for $emptyset$ and $mathbbR$. And to your third question, $(infty, a)$ is not open in the topology, and so you don't need to write it as a union of the basis elements.
– Dabbler
Sep 10 at 21:43
My previous comment also answers your other two questions. Since $B cup emptyset, mathbbR$ is a topological space, you can restrict your attention to sets of the form $A = (a, infty)$ when discussing interior and closure operations; the remaining open sets are trivial. The answer to your second question is "yes", if you add cases for $emptyset$ and $mathbbR$. And to your third question, $(infty, a)$ is not open in the topology, and so you don't need to write it as a union of the basis elements.
– Dabbler
Sep 10 at 21:43
add a comment |Â
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There are several ways to understand the closure operation in this topology. Let's examine one.
Let $p in A$. If $x le p$, let $O$ be any basic set that contains $x$, so $O = (a,infty)$ with $x in (a,infty)$, or $a < x$. As $a < x le p$, $p in O cap A$ and so $x in overlineA$, because every basic open set containing it intersects $A$.
So if $A$ is any set, the set $A^downarrow := x in mathbbR: exists p in A: x le p subseteq overlineA$.
Now if $x notin A^downarrow$, this means that $forall p in A: x > p$, or $x$ is an upperbound for $A$. And in that case, (for any $p in A$) $(p ,infty)$ is an open subset of $x$ in this topology that misses $A$ entirely, so that $x notin overlineA$.
This shows that $A^downarrow = overlineA$, for $A subseteq mathbbR$.
In particular $overline(a,infty) = mathbbR$ for all $a in mathbbR$, so I don't agree with the linked solution that claims $overline(a, infty)$ just adds the point $a$ to $(a,infty)$. All open subsets are "right facing" so it's quite clear that $[a,infty) =a cup (a,infty)$ is not even closed in this topology (its complement is "left facing").
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There are several ways to understand the closure operation in this topology. Let's examine one.
Let $p in A$. If $x le p$, let $O$ be any basic set that contains $x$, so $O = (a,infty)$ with $x in (a,infty)$, or $a < x$. As $a < x le p$, $p in O cap A$ and so $x in overlineA$, because every basic open set containing it intersects $A$.
So if $A$ is any set, the set $A^downarrow := x in mathbbR: exists p in A: x le p subseteq overlineA$.
Now if $x notin A^downarrow$, this means that $forall p in A: x > p$, or $x$ is an upperbound for $A$. And in that case, (for any $p in A$) $(p ,infty)$ is an open subset of $x$ in this topology that misses $A$ entirely, so that $x notin overlineA$.
This shows that $A^downarrow = overlineA$, for $A subseteq mathbbR$.
In particular $overline(a,infty) = mathbbR$ for all $a in mathbbR$, so I don't agree with the linked solution that claims $overline(a, infty)$ just adds the point $a$ to $(a,infty)$. All open subsets are "right facing" so it's quite clear that $[a,infty) =a cup (a,infty)$ is not even closed in this topology (its complement is "left facing").
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
add a comment |Â
up vote
1
down vote
accepted
There are several ways to understand the closure operation in this topology. Let's examine one.
Let $p in A$. If $x le p$, let $O$ be any basic set that contains $x$, so $O = (a,infty)$ with $x in (a,infty)$, or $a < x$. As $a < x le p$, $p in O cap A$ and so $x in overlineA$, because every basic open set containing it intersects $A$.
So if $A$ is any set, the set $A^downarrow := x in mathbbR: exists p in A: x le p subseteq overlineA$.
Now if $x notin A^downarrow$, this means that $forall p in A: x > p$, or $x$ is an upperbound for $A$. And in that case, (for any $p in A$) $(p ,infty)$ is an open subset of $x$ in this topology that misses $A$ entirely, so that $x notin overlineA$.
This shows that $A^downarrow = overlineA$, for $A subseteq mathbbR$.
In particular $overline(a,infty) = mathbbR$ for all $a in mathbbR$, so I don't agree with the linked solution that claims $overline(a, infty)$ just adds the point $a$ to $(a,infty)$. All open subsets are "right facing" so it's quite clear that $[a,infty) =a cup (a,infty)$ is not even closed in this topology (its complement is "left facing").
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There are several ways to understand the closure operation in this topology. Let's examine one.
Let $p in A$. If $x le p$, let $O$ be any basic set that contains $x$, so $O = (a,infty)$ with $x in (a,infty)$, or $a < x$. As $a < x le p$, $p in O cap A$ and so $x in overlineA$, because every basic open set containing it intersects $A$.
So if $A$ is any set, the set $A^downarrow := x in mathbbR: exists p in A: x le p subseteq overlineA$.
Now if $x notin A^downarrow$, this means that $forall p in A: x > p$, or $x$ is an upperbound for $A$. And in that case, (for any $p in A$) $(p ,infty)$ is an open subset of $x$ in this topology that misses $A$ entirely, so that $x notin overlineA$.
This shows that $A^downarrow = overlineA$, for $A subseteq mathbbR$.
In particular $overline(a,infty) = mathbbR$ for all $a in mathbbR$, so I don't agree with the linked solution that claims $overline(a, infty)$ just adds the point $a$ to $(a,infty)$. All open subsets are "right facing" so it's quite clear that $[a,infty) =a cup (a,infty)$ is not even closed in this topology (its complement is "left facing").
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
There are several ways to understand the closure operation in this topology. Let's examine one.
Let $p in A$. If $x le p$, let $O$ be any basic set that contains $x$, so $O = (a,infty)$ with $x in (a,infty)$, or $a < x$. As $a < x le p$, $p in O cap A$ and so $x in overlineA$, because every basic open set containing it intersects $A$.
So if $A$ is any set, the set $A^downarrow := x in mathbbR: exists p in A: x le p subseteq overlineA$.
Now if $x notin A^downarrow$, this means that $forall p in A: x > p$, or $x$ is an upperbound for $A$. And in that case, (for any $p in A$) $(p ,infty)$ is an open subset of $x$ in this topology that misses $A$ entirely, so that $x notin overlineA$.
This shows that $A^downarrow = overlineA$, for $A subseteq mathbbR$.
In particular $overline(a,infty) = mathbbR$ for all $a in mathbbR$, so I don't agree with the linked solution that claims $overline(a, infty)$ just adds the point $a$ to $(a,infty)$. All open subsets are "right facing" so it's quite clear that $[a,infty) =a cup (a,infty)$ is not even closed in this topology (its complement is "left facing").
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
edited Sep 10 at 21:22
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
answered Sep 10 at 21:17
Henno Brandsma
94.1k343102
94.1k343102
add a comment |Â
add a comment |Â
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1
Per Henno Brandsma's answer below, here's another way to see that $overline(a,infty)$ is the entire real line. Check that $B cup emptyset, mathbbR$ is a topological space. So the only closed sets in the topology are $emptyset$, $mathbbR$, and complements of elements of $B$, The only closed set containing $(a, infty)$, therefore, is $mathbbR$.
– Dabbler
Sep 10 at 21:30
1
My previous comment also answers your other two questions. Since $B cup emptyset, mathbbR$ is a topological space, you can restrict your attention to sets of the form $A = (a, infty)$ when discussing interior and closure operations; the remaining open sets are trivial. The answer to your second question is "yes", if you add cases for $emptyset$ and $mathbbR$. And to your third question, $(infty, a)$ is not open in the topology, and so you don't need to write it as a union of the basis elements.
– Dabbler
Sep 10 at 21:43