Mixing
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Left Semigroups and Compact sets
$begingroup$
I have fallen down the rabbit hole of reading old papers. I (believe that I) have read the following claim:
Let $(S,cdot)$ be a semigroup and assume that $S$ is also a topological space such that for each $b in S$, the map $- cdot b: S to S$ is a continuous map. Let $A,B subset S$ such that $A,B$ are compact. Then, $A cdot B$ is compact.
I might just be getting old and rusty, but I can't seem to show this. Do I need to assume that $cdot$ is jointly continuous? In particular, that the maps $a cdot -$ are continuous?
Edit: I've added the [model-theory] tag since I assume someone here has had thoughts on this before in relation to the Ellis Semigroup.
general-topology model-theory semigroups
asked Jan 27 at 4:41
KyleKyle
5,28721334
$endgroup$
add a comment |
$begingroup$
I have fallen down the rabbit hole of reading old papers. I (believe that I) have read the following claim:
Let $(S,cdot)$ be a semigroup and assume that $S$ is also a topological space such that for each $b in S$, the map $- cdot b: S to S$ is a continuous map. Let $A,B subset S$ such that $A,B$ are compact. Then, $A cdot B$ is compact.
I might just be getting old and rusty, but I can't seem to show this. Do I need to assume that $cdot$ is jointly continuous? In particular, that the maps $a cdot -$ are continuous?
Edit: I've added the [model-theory] tag since I assume someone here has had thoughts on this before in relation to the Ellis Semigroup.
general-topology model-theory semigroups
asked Jan 27 at 4:41
KyleKyle
5,28721334
$endgroup$
add a comment |
$begingroup$
I have fallen down the rabbit hole of reading old papers. I (believe that I) have read the following claim:
Let $(S,cdot)$ be a semigroup and assume that $S$ is also a topological space such that for each $b in S$, the map $- cdot b: S to S$ is a continuous map. Let $A,B subset S$ such that $A,B$ are compact. Then, $A cdot B$ is compact.
I might just be getting old and rusty, but I can't seem to show this. Do I need to assume that $cdot$ is jointly continuous? In particular, that the maps $a cdot -$ are continuous?
Edit: I've added the [model-theory] tag since I assume someone here has had thoughts on this before in relation to the Ellis Semigroup.
general-topology model-theory semigroups
asked Jan 27 at 4:41
KyleKyle
5,28721334
$endgroup$
I have fallen down the rabbit hole of reading old papers. I (believe that I) have read the following claim:
Let $(S,cdot)$ be a semigroup and assume that $S$ is also a topological space such that for each $b in S$, the map $- cdot b: S to S$ is a continuous map. Let $A,B subset S$ such that $A,B$ are compact. Then, $A cdot B$ is compact.
I might just be getting old and rusty, but I can't seem to show this. Do I need to assume that $cdot$ is jointly continuous? In particular, that the maps $a cdot -$ are continuous?
Edit: I've added the [model-theory] tag since I assume someone here has had thoughts on this before in relation to the Ellis Semigroup.
general-topology model-theory semigroups
general-topology model-theory semigroups
asked Jan 27 at 4:41
KyleKyle
5,28721334
asked Jan 27 at 4:41
KyleKyle
5,28721334
edited Jan 27 at 4:51
Kyle
asked Jan 27 at 4:41
KyleKyle
5,28721334
asked Jan 27 at 4:41
KyleKyle
5,28721334
asked Jan 27 at 4:41
KyleKyle
5,28721334
5,28721334
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
This isn't true under the hypotheses you gave. For example, let $S$ be the real numbers with their usual topology, and define $acdot b = 0$ if $b$ is an integer and $acdot b = ab$ (usual multiplication) otherwise. Then $(-cdot b)$ is continuous for any fixed $b$, but $[0,1]cdot[0,1]=[0,1)$.
Edit: As J.-E. Pin points out in the comments, the example I gave above is not actually a semigroup. Here's an example that actually works:
Let $S = mathbb{R}cup {infty}$ be the one-point compactification of $mathbb{R}$ (the circle), with a semigroup operation defined by: $$acdot b = begin{cases} 0& text{if }b = inftytext{ or }b = 0\ ab & text{if }ain mathbb{R} text{ and }bin mathbb{R}^times\ infty &text{if }a = inftytext{ and }bin mathbb{R}^times. end{cases}$$
For any $b$, $(-cdot b) colon Sto S$ is continuous. This map is constant if $b = infty$ or $b = 0$, and otherwise it is the "multiply by $b$" map $mathbb{R}to mathbb{R}$ extended continuously to a map $Sto S$. And $cdot$ is associative: it is easy to check that $$(acdot b)cdot c = acdot (bcdot c) = begin{cases} abc & text{if }a,b,cin mathbb{R}^times\
infty & text{if }a = infty text{ and }b,cin mathbb{R}^times\
0 & text{otherwise}.end{cases}$$
The sets ${1}$ and $S$ are compact, but ${1}cdot S = mathbb{R}$, which is not compact.
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
2
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
|
show 1 more comment
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1 Answer
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$begingroup$
This isn't true under the hypotheses you gave. For example, let $S$ be the real numbers with their usual topology, and define $acdot b = 0$ if $b$ is an integer and $acdot b = ab$ (usual multiplication) otherwise. Then $(-cdot b)$ is continuous for any fixed $b$, but $[0,1]cdot[0,1]=[0,1)$.
Edit: As J.-E. Pin points out in the comments, the example I gave above is not actually a semigroup. Here's an example that actually works:
Let $S = mathbb{R}cup {infty}$ be the one-point compactification of $mathbb{R}$ (the circle), with a semigroup operation defined by: $$acdot b = begin{cases} 0& text{if }b = inftytext{ or }b = 0\ ab & text{if }ain mathbb{R} text{ and }bin mathbb{R}^times\ infty &text{if }a = inftytext{ and }bin mathbb{R}^times. end{cases}$$
For any $b$, $(-cdot b) colon Sto S$ is continuous. This map is constant if $b = infty$ or $b = 0$, and otherwise it is the "multiply by $b$" map $mathbb{R}to mathbb{R}$ extended continuously to a map $Sto S$. And $cdot$ is associative: it is easy to check that $$(acdot b)cdot c = acdot (bcdot c) = begin{cases} abc & text{if }a,b,cin mathbb{R}^times\
infty & text{if }a = infty text{ and }b,cin mathbb{R}^times\
0 & text{otherwise}.end{cases}$$
The sets ${1}$ and $S$ are compact, but ${1}cdot S = mathbb{R}$, which is not compact.
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
2
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
|
show 1 more comment
$begingroup$
This isn't true under the hypotheses you gave. For example, let $S$ be the real numbers with their usual topology, and define $acdot b = 0$ if $b$ is an integer and $acdot b = ab$ (usual multiplication) otherwise. Then $(-cdot b)$ is continuous for any fixed $b$, but $[0,1]cdot[0,1]=[0,1)$.
Edit: As J.-E. Pin points out in the comments, the example I gave above is not actually a semigroup. Here's an example that actually works:
Let $S = mathbb{R}cup {infty}$ be the one-point compactification of $mathbb{R}$ (the circle), with a semigroup operation defined by: $$acdot b = begin{cases} 0& text{if }b = inftytext{ or }b = 0\ ab & text{if }ain mathbb{R} text{ and }bin mathbb{R}^times\ infty &text{if }a = inftytext{ and }bin mathbb{R}^times. end{cases}$$
For any $b$, $(-cdot b) colon Sto S$ is continuous. This map is constant if $b = infty$ or $b = 0$, and otherwise it is the "multiply by $b$" map $mathbb{R}to mathbb{R}$ extended continuously to a map $Sto S$. And $cdot$ is associative: it is easy to check that $$(acdot b)cdot c = acdot (bcdot c) = begin{cases} abc & text{if }a,b,cin mathbb{R}^times\
infty & text{if }a = infty text{ and }b,cin mathbb{R}^times\
0 & text{otherwise}.end{cases}$$
The sets ${1}$ and $S$ are compact, but ${1}cdot S = mathbb{R}$, which is not compact.
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
2
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
|
show 1 more comment
$begingroup$
This isn't true under the hypotheses you gave. For example, let $S$ be the real numbers with their usual topology, and define $acdot b = 0$ if $b$ is an integer and $acdot b = ab$ (usual multiplication) otherwise. Then $(-cdot b)$ is continuous for any fixed $b$, but $[0,1]cdot[0,1]=[0,1)$.
Edit: As J.-E. Pin points out in the comments, the example I gave above is not actually a semigroup. Here's an example that actually works:
Let $S = mathbb{R}cup {infty}$ be the one-point compactification of $mathbb{R}$ (the circle), with a semigroup operation defined by: $$acdot b = begin{cases} 0& text{if }b = inftytext{ or }b = 0\ ab & text{if }ain mathbb{R} text{ and }bin mathbb{R}^times\ infty &text{if }a = inftytext{ and }bin mathbb{R}^times. end{cases}$$
For any $b$, $(-cdot b) colon Sto S$ is continuous. This map is constant if $b = infty$ or $b = 0$, and otherwise it is the "multiply by $b$" map $mathbb{R}to mathbb{R}$ extended continuously to a map $Sto S$. And $cdot$ is associative: it is easy to check that $$(acdot b)cdot c = acdot (bcdot c) = begin{cases} abc & text{if }a,b,cin mathbb{R}^times\
infty & text{if }a = infty text{ and }b,cin mathbb{R}^times\
0 & text{otherwise}.end{cases}$$
The sets ${1}$ and $S$ are compact, but ${1}cdot S = mathbb{R}$, which is not compact.
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
$endgroup$
This isn't true under the hypotheses you gave. For example, let $S$ be the real numbers with their usual topology, and define $acdot b = 0$ if $b$ is an integer and $acdot b = ab$ (usual multiplication) otherwise. Then $(-cdot b)$ is continuous for any fixed $b$, but $[0,1]cdot[0,1]=[0,1)$.
Edit: As J.-E. Pin points out in the comments, the example I gave above is not actually a semigroup. Here's an example that actually works:
Let $S = mathbb{R}cup {infty}$ be the one-point compactification of $mathbb{R}$ (the circle), with a semigroup operation defined by: $$acdot b = begin{cases} 0& text{if }b = inftytext{ or }b = 0\ ab & text{if }ain mathbb{R} text{ and }bin mathbb{R}^times\ infty &text{if }a = inftytext{ and }bin mathbb{R}^times. end{cases}$$
For any $b$, $(-cdot b) colon Sto S$ is continuous. This map is constant if $b = infty$ or $b = 0$, and otherwise it is the "multiply by $b$" map $mathbb{R}to mathbb{R}$ extended continuously to a map $Sto S$. And $cdot$ is associative: it is easy to check that $$(acdot b)cdot c = acdot (bcdot c) = begin{cases} abc & text{if }a,b,cin mathbb{R}^times\
infty & text{if }a = infty text{ and }b,cin mathbb{R}^times\
0 & text{otherwise}.end{cases}$$
The sets ${1}$ and $S$ are compact, but ${1}cdot S = mathbb{R}$, which is not compact.
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
edited Feb 12 at 18:49
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
answered Jan 27 at 15:33
Alex KruckmanAlex Kruckman
28.1k32658
28.1k32658
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
2
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
|
show 1 more comment
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
2
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Nice example; my books were suggestive (in exercises) that it would even fail for quasitopological groups and semitopological groups as well (which are a bit stronger). I couldn't find easy examples for those either.
$endgroup$
– Henno Brandsma
Jan 27 at 16:08
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
$begingroup$
Thanks Alex. That is a very good example.
$endgroup$
– Kyle
Jan 27 at 16:27
2
2
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
I am afraid your example is not a semigroup. For instance, $(1cdot{2 over 3})cdot{3over 2} = {2 over 3}cdot{3over 2} = 1$ but $1cdot({2 over 3}cdot{3over 2}) = 1cdot1 = 0$.
$endgroup$
– J.-E. Pin
Feb 12 at 11:46
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Indeed! Do you have a different example?
$endgroup$
– Alex Kruckman
Feb 12 at 13:20
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
$begingroup$
@J.-E.Pin Ah, I found an example that works. I'll edit it in later today.
$endgroup$
– Alex Kruckman
Feb 12 at 17:27
|
show 1 more comment
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