Mixing
Is sum of two closed set is closed
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
If A, B are closed subsets of $[0,∞)$, then
$$A + B = ,x + y mid x ∈ A,, y ∈ B,$$
is closed in $ [0,infty)$.
If A ,B are closed sets in $mathbb R$ , then i know counterexamples but considering only non negative reals ,i can't able to get a counterexample which doesn't implies it must be true.
So i want to prove or disprove , I know it is not true in genral but under this particular situation i want to know it is TRUE /FALSE
real-analysis general-topology
edited 1 hour ago
Hagen von Eitzen
272k21264492
asked 1 hour ago
Cloud JR
604415
add a comment |Â
up vote
1
down vote
favorite
If A, B are closed subsets of $[0,∞)$, then
$$A + B = ,x + y mid x ∈ A,, y ∈ B,$$
is closed in $ [0,infty)$.
If A ,B are closed sets in $mathbb R$ , then i know counterexamples but considering only non negative reals ,i can't able to get a counterexample which doesn't implies it must be true.
So i want to prove or disprove , I know it is not true in genral but under this particular situation i want to know it is TRUE /FALSE
real-analysis general-topology
edited 1 hour ago
Hagen von Eitzen
272k21264492
asked 1 hour ago
Cloud JR
604415
In case you didn't see my reply to your comment beneath my (now deleted) answer, you were quite right, my answer was quite wrong. Sorry for the noise.
– Barry Cipra
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If A, B are closed subsets of $[0,∞)$, then
$$A + B = ,x + y mid x ∈ A,, y ∈ B,$$
is closed in $ [0,infty)$.
If A ,B are closed sets in $mathbb R$ , then i know counterexamples but considering only non negative reals ,i can't able to get a counterexample which doesn't implies it must be true.
So i want to prove or disprove , I know it is not true in genral but under this particular situation i want to know it is TRUE /FALSE
real-analysis general-topology
edited 1 hour ago
Hagen von Eitzen
272k21264492
asked 1 hour ago
Cloud JR
604415
If A, B are closed subsets of $[0,∞)$, then
$$A + B = ,x + y mid x ∈ A,, y ∈ B,$$
is closed in $ [0,infty)$.
If A ,B are closed sets in $mathbb R$ , then i know counterexamples but considering only non negative reals ,i can't able to get a counterexample which doesn't implies it must be true.
So i want to prove or disprove , I know it is not true in genral but under this particular situation i want to know it is TRUE /FALSE
real-analysis general-topology
real-analysis general-topology
edited 1 hour ago
Hagen von Eitzen
272k21264492
asked 1 hour ago
Cloud JR
604415
edited 1 hour ago
Hagen von Eitzen
272k21264492
asked 1 hour ago
Cloud JR
604415
edited 1 hour ago
Hagen von Eitzen
272k21264492
edited 1 hour ago
Hagen von Eitzen
272k21264492
edited 1 hour ago
Hagen von Eitzen
272k21264492
272k21264492
asked 1 hour ago
Cloud JR
604415
asked 1 hour ago
Cloud JR
604415
asked 1 hour ago
Cloud JR
604415
604415
In case you didn't see my reply to your comment beneath my (now deleted) answer, you were quite right, my answer was quite wrong. Sorry for the noise.
– Barry Cipra
1 hour ago
add a comment |Â
In case you didn't see my reply to your comment beneath my (now deleted) answer, you were quite right, my answer was quite wrong. Sorry for the noise.
– Barry Cipra
1 hour ago
In case you didn't see my reply to your comment beneath my (now deleted) answer, you were quite right, my answer was quite wrong. Sorry for the noise.
– Barry Cipra
1 hour ago
In case you didn't see my reply to your comment beneath my (now deleted) answer, you were quite right, my answer was quite wrong. Sorry for the noise.
– Barry Cipra
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Take a sequence $(x_n)subseteq A+B$ convergent to $xin [0,infty)$. We want to show that $xin A+B$.
It follows that $x_n=a_n+b_n$ for some sequences $(a_n)subseteq A$ and $(b_n)subseteq B$. Since $A,B$ are subsets of $[0,infty)$ then $$0leq a_nleq x_n$$
$$0leq b_nleq x_n$$
In particular, since $x_n$ is bounded then so are $a_n$ and $b_n$ (this is exactly the place where the proof would fail for whole $mathbbR$ case). Therefore they have convergent subsequences, say $a_n_k$ and $b_n_k$ (we can choose indexes in such a way that they coincide). It follows that
$$x=lim_n=0^infty x_n=lim_k=0^infty x_n_k= lim_k=0^inftybig(a_n_k+b_n_kbig)=lim_k=0^inftya_n_k+lim_k=0^inftyb_n_kin A+B$$
the last "$in$" because $A,B$ are closed.
answered 1 hour ago
freakish
9,6561524
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
add a comment |Â
up vote
1
down vote
If $A$ and $B$ are compact, $A + B$ is compact because it is the image of the compact set $A times B$ under the continuous function $(a,b) mapsto a+b$.
If $A$ and $B$ are closed subsets of $[0,infty)$, then for all $N > 0$ we have
$$(A + B) cap [0, N] = ((A cap [0,N]) + (B cap [0,N])) cap [0,N]$$ which is closed, therefore $A + B$ is closed.
answered 1 hour ago
Robert Israel
312k23204451
add a comment |Â
up vote
0
down vote
Consider the compactification $X = [0,+infty]$. We can extend the addition function to $X times X$ by defining $+infty + x = x + +infty = +infty$ for all $x in X$, and one can check this is still continuous as a map $p$ from $X times X to X$. Then, if $A,B subseteq [0,+infty)$ are closed, then their closures in $X$ are either $A$ resp $B$ or $A cup +infty$ or $B cup +infty$ respectively, depending on whether $A$ and $B$ were bounded or not.
In either case $p[overlineA times overlineB]$ is compact hence closed in $X$ and so $A + B = p[overlineA times overlineB] cap [0,+infty)$ is also closed in $[0,+infty)$.
answered 16 mins ago
Henno Brandsma
99.4k344107
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Take a sequence $(x_n)subseteq A+B$ convergent to $xin [0,infty)$. We want to show that $xin A+B$.
It follows that $x_n=a_n+b_n$ for some sequences $(a_n)subseteq A$ and $(b_n)subseteq B$. Since $A,B$ are subsets of $[0,infty)$ then $$0leq a_nleq x_n$$
$$0leq b_nleq x_n$$
In particular, since $x_n$ is bounded then so are $a_n$ and $b_n$ (this is exactly the place where the proof would fail for whole $mathbbR$ case). Therefore they have convergent subsequences, say $a_n_k$ and $b_n_k$ (we can choose indexes in such a way that they coincide). It follows that
$$x=lim_n=0^infty x_n=lim_k=0^infty x_n_k= lim_k=0^inftybig(a_n_k+b_n_kbig)=lim_k=0^inftya_n_k+lim_k=0^inftyb_n_kin A+B$$
the last "$in$" because $A,B$ are closed.
answered 1 hour ago
freakish
9,6561524
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
add a comment |Â
up vote
4
down vote
accepted
Take a sequence $(x_n)subseteq A+B$ convergent to $xin [0,infty)$. We want to show that $xin A+B$.
It follows that $x_n=a_n+b_n$ for some sequences $(a_n)subseteq A$ and $(b_n)subseteq B$. Since $A,B$ are subsets of $[0,infty)$ then $$0leq a_nleq x_n$$
$$0leq b_nleq x_n$$
In particular, since $x_n$ is bounded then so are $a_n$ and $b_n$ (this is exactly the place where the proof would fail for whole $mathbbR$ case). Therefore they have convergent subsequences, say $a_n_k$ and $b_n_k$ (we can choose indexes in such a way that they coincide). It follows that
$$x=lim_n=0^infty x_n=lim_k=0^infty x_n_k= lim_k=0^inftybig(a_n_k+b_n_kbig)=lim_k=0^inftya_n_k+lim_k=0^inftyb_n_kin A+B$$
the last "$in$" because $A,B$ are closed.
answered 1 hour ago
freakish
9,6561524
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Take a sequence $(x_n)subseteq A+B$ convergent to $xin [0,infty)$. We want to show that $xin A+B$.
It follows that $x_n=a_n+b_n$ for some sequences $(a_n)subseteq A$ and $(b_n)subseteq B$. Since $A,B$ are subsets of $[0,infty)$ then $$0leq a_nleq x_n$$
$$0leq b_nleq x_n$$
In particular, since $x_n$ is bounded then so are $a_n$ and $b_n$ (this is exactly the place where the proof would fail for whole $mathbbR$ case). Therefore they have convergent subsequences, say $a_n_k$ and $b_n_k$ (we can choose indexes in such a way that they coincide). It follows that
$$x=lim_n=0^infty x_n=lim_k=0^infty x_n_k= lim_k=0^inftybig(a_n_k+b_n_kbig)=lim_k=0^inftya_n_k+lim_k=0^inftyb_n_kin A+B$$
the last "$in$" because $A,B$ are closed.
answered 1 hour ago
freakish
9,6561524
Take a sequence $(x_n)subseteq A+B$ convergent to $xin [0,infty)$. We want to show that $xin A+B$.
It follows that $x_n=a_n+b_n$ for some sequences $(a_n)subseteq A$ and $(b_n)subseteq B$. Since $A,B$ are subsets of $[0,infty)$ then $$0leq a_nleq x_n$$
$$0leq b_nleq x_n$$
In particular, since $x_n$ is bounded then so are $a_n$ and $b_n$ (this is exactly the place where the proof would fail for whole $mathbbR$ case). Therefore they have convergent subsequences, say $a_n_k$ and $b_n_k$ (we can choose indexes in such a way that they coincide). It follows that
$$x=lim_n=0^infty x_n=lim_k=0^infty x_n_k= lim_k=0^inftybig(a_n_k+b_n_kbig)=lim_k=0^inftya_n_k+lim_k=0^inftyb_n_kin A+B$$
the last "$in$" because $A,B$ are closed.
answered 1 hour ago
freakish
9,6561524
edited 27 mins ago
answered 1 hour ago
freakish
9,6561524
answered 1 hour ago
freakish
9,6561524
answered 1 hour ago
freakish
9,6561524
9,6561524
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
add a comment |Â
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
We don't know anything about uniqueness of subsequential limit of $a_n_k$ and $b_n_k $. Is it a problem? To uniqueness of x?
– Cloud JR
1 hour ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
@CloudJR You mean that these limits depend on the choice of subsequences? Yes, but it doesn't matter. The only important thing is that they belong to $A,B$ respectively and their sum is $x$. Note that the choice of whole $(a_n), (b_n)$ sequences is not unique as well. In particular if $A=B=[0,infty)$ then there's a lot of different choices.
– freakish
30 mins ago
add a comment |Â
up vote
1
down vote
If $A$ and $B$ are compact, $A + B$ is compact because it is the image of the compact set $A times B$ under the continuous function $(a,b) mapsto a+b$.
If $A$ and $B$ are closed subsets of $[0,infty)$, then for all $N > 0$ we have
$$(A + B) cap [0, N] = ((A cap [0,N]) + (B cap [0,N])) cap [0,N]$$ which is closed, therefore $A + B$ is closed.
answered 1 hour ago
Robert Israel
312k23204451
add a comment |Â
up vote
1
down vote
If $A$ and $B$ are compact, $A + B$ is compact because it is the image of the compact set $A times B$ under the continuous function $(a,b) mapsto a+b$.
If $A$ and $B$ are closed subsets of $[0,infty)$, then for all $N > 0$ we have
$$(A + B) cap [0, N] = ((A cap [0,N]) + (B cap [0,N])) cap [0,N]$$ which is closed, therefore $A + B$ is closed.
answered 1 hour ago
Robert Israel
312k23204451
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $A$ and $B$ are compact, $A + B$ is compact because it is the image of the compact set $A times B$ under the continuous function $(a,b) mapsto a+b$.
If $A$ and $B$ are closed subsets of $[0,infty)$, then for all $N > 0$ we have
$$(A + B) cap [0, N] = ((A cap [0,N]) + (B cap [0,N])) cap [0,N]$$ which is closed, therefore $A + B$ is closed.
answered 1 hour ago
Robert Israel
312k23204451
If $A$ and $B$ are compact, $A + B$ is compact because it is the image of the compact set $A times B$ under the continuous function $(a,b) mapsto a+b$.
If $A$ and $B$ are closed subsets of $[0,infty)$, then for all $N > 0$ we have
$$(A + B) cap [0, N] = ((A cap [0,N]) + (B cap [0,N])) cap [0,N]$$ which is closed, therefore $A + B$ is closed.
answered 1 hour ago
Robert Israel
312k23204451
answered 1 hour ago
Robert Israel
312k23204451
answered 1 hour ago
Robert Israel
312k23204451
answered 1 hour ago
Robert Israel
312k23204451
312k23204451
add a comment |Â
add a comment |Â
up vote
0
down vote
Consider the compactification $X = [0,+infty]$. We can extend the addition function to $X times X$ by defining $+infty + x = x + +infty = +infty$ for all $x in X$, and one can check this is still continuous as a map $p$ from $X times X to X$. Then, if $A,B subseteq [0,+infty)$ are closed, then their closures in $X$ are either $A$ resp $B$ or $A cup +infty$ or $B cup +infty$ respectively, depending on whether $A$ and $B$ were bounded or not.
In either case $p[overlineA times overlineB]$ is compact hence closed in $X$ and so $A + B = p[overlineA times overlineB] cap [0,+infty)$ is also closed in $[0,+infty)$.
answered 16 mins ago
Henno Brandsma
99.4k344107
add a comment |Â
up vote
0
down vote
Consider the compactification $X = [0,+infty]$. We can extend the addition function to $X times X$ by defining $+infty + x = x + +infty = +infty$ for all $x in X$, and one can check this is still continuous as a map $p$ from $X times X to X$. Then, if $A,B subseteq [0,+infty)$ are closed, then their closures in $X$ are either $A$ resp $B$ or $A cup +infty$ or $B cup +infty$ respectively, depending on whether $A$ and $B$ were bounded or not.
In either case $p[overlineA times overlineB]$ is compact hence closed in $X$ and so $A + B = p[overlineA times overlineB] cap [0,+infty)$ is also closed in $[0,+infty)$.
answered 16 mins ago
Henno Brandsma
99.4k344107
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider the compactification $X = [0,+infty]$. We can extend the addition function to $X times X$ by defining $+infty + x = x + +infty = +infty$ for all $x in X$, and one can check this is still continuous as a map $p$ from $X times X to X$. Then, if $A,B subseteq [0,+infty)$ are closed, then their closures in $X$ are either $A$ resp $B$ or $A cup +infty$ or $B cup +infty$ respectively, depending on whether $A$ and $B$ were bounded or not.
In either case $p[overlineA times overlineB]$ is compact hence closed in $X$ and so $A + B = p[overlineA times overlineB] cap [0,+infty)$ is also closed in $[0,+infty)$.
answered 16 mins ago
Henno Brandsma
99.4k344107
Consider the compactification $X = [0,+infty]$. We can extend the addition function to $X times X$ by defining $+infty + x = x + +infty = +infty$ for all $x in X$, and one can check this is still continuous as a map $p$ from $X times X to X$. Then, if $A,B subseteq [0,+infty)$ are closed, then their closures in $X$ are either $A$ resp $B$ or $A cup +infty$ or $B cup +infty$ respectively, depending on whether $A$ and $B$ were bounded or not.
In either case $p[overlineA times overlineB]$ is compact hence closed in $X$ and so $A + B = p[overlineA times overlineB] cap [0,+infty)$ is also closed in $[0,+infty)$.
answered 16 mins ago
Henno Brandsma
99.4k344107
answered 16 mins ago
Henno Brandsma
99.4k344107
answered 16 mins ago
Henno Brandsma
99.4k344107
answered 16 mins ago
Henno Brandsma
99.4k344107
99.4k344107
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%2f2985787%2fis-sum-of-two-closed-set-is-closed%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
In case you didn't see my reply to your comment beneath my (now deleted) answer, you were quite right, my answer was quite wrong. Sorry for the noise.
– Barry Cipra
1 hour ago