Mixing
Discreteness“+” Compactness= Finiteness: What does this really mean?
$begingroup$
In my journey of self-studying topology, I have always come here to gain some insights on topics that really didn't click for me, and it's marvelous to have such a place where you can come back when you feel a little bit "scared".
That's what happened when I was introduced to the concept of compactness.
Lost in the interpretation of what a finite subcover could mean, I came here in search for intuition, and I have to say, as always, that I was really satisfied for what I got back.
On this popular post about the topic, answers are different and very interesting, so it's not my intent to open again a similiar discussion.
The aim of this post is to ask for a more exhaustive presentation of the insight given by the most-voted answer.
I loved the idea depicted by the author with a story, about finiteness encapsulating the notions of compactness and discreteness, but I was a little bit disappointed about it ending without a demonstration/examples of this fact.
Being someone who learns by categorizing, I was enthusiastic about the idea of thinking about finiteness as unifying the concepts of compactness and discreteness, but I remained with this intuition without being able to realize why this was true or made sense.
So my hope is that through this post I can permit someone to delve deeper on this topic. What are the "compactness attributes" of finiteness? What are the "discreteness attributes" of finiteness?
general-topology analysis compactness
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
$endgroup$
add a comment |
$begingroup$
In my journey of self-studying topology, I have always come here to gain some insights on topics that really didn't click for me, and it's marvelous to have such a place where you can come back when you feel a little bit "scared".
That's what happened when I was introduced to the concept of compactness.
Lost in the interpretation of what a finite subcover could mean, I came here in search for intuition, and I have to say, as always, that I was really satisfied for what I got back.
On this popular post about the topic, answers are different and very interesting, so it's not my intent to open again a similiar discussion.
The aim of this post is to ask for a more exhaustive presentation of the insight given by the most-voted answer.
I loved the idea depicted by the author with a story, about finiteness encapsulating the notions of compactness and discreteness, but I was a little bit disappointed about it ending without a demonstration/examples of this fact.
Being someone who learns by categorizing, I was enthusiastic about the idea of thinking about finiteness as unifying the concepts of compactness and discreteness, but I remained with this intuition without being able to realize why this was true or made sense.
So my hope is that through this post I can permit someone to delve deeper on this topic. What are the "compactness attributes" of finiteness? What are the "discreteness attributes" of finiteness?
general-topology analysis compactness
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
$endgroup$
2
$begingroup$
I recommend proving many examples are not compact to see what happens when you can’t find any finite subcover that works in order to sharpen intuition about the concept. E.g. Try to show $mathbb{N}$ and $(0,1]$ are not compact in $mathbb{R}$, though maybe you’ve done this already and are still grasping for the idea. Just a suggestion.
$endgroup$
– LoveTooNap29
Jan 29 at 16:55
$begingroup$
I read this article which talks about it, I'll post it maybe it will be useful to someone: blogs.scientificamerican.com/roots-of-unity/…
$endgroup$
– Gabriele Scarlatti
Jan 29 at 16:57
add a comment |
$begingroup$
In my journey of self-studying topology, I have always come here to gain some insights on topics that really didn't click for me, and it's marvelous to have such a place where you can come back when you feel a little bit "scared".
That's what happened when I was introduced to the concept of compactness.
Lost in the interpretation of what a finite subcover could mean, I came here in search for intuition, and I have to say, as always, that I was really satisfied for what I got back.
On this popular post about the topic, answers are different and very interesting, so it's not my intent to open again a similiar discussion.
The aim of this post is to ask for a more exhaustive presentation of the insight given by the most-voted answer.
I loved the idea depicted by the author with a story, about finiteness encapsulating the notions of compactness and discreteness, but I was a little bit disappointed about it ending without a demonstration/examples of this fact.
Being someone who learns by categorizing, I was enthusiastic about the idea of thinking about finiteness as unifying the concepts of compactness and discreteness, but I remained with this intuition without being able to realize why this was true or made sense.
So my hope is that through this post I can permit someone to delve deeper on this topic. What are the "compactness attributes" of finiteness? What are the "discreteness attributes" of finiteness?
general-topology analysis compactness
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
$endgroup$
In my journey of self-studying topology, I have always come here to gain some insights on topics that really didn't click for me, and it's marvelous to have such a place where you can come back when you feel a little bit "scared".
That's what happened when I was introduced to the concept of compactness.
Lost in the interpretation of what a finite subcover could mean, I came here in search for intuition, and I have to say, as always, that I was really satisfied for what I got back.
On this popular post about the topic, answers are different and very interesting, so it's not my intent to open again a similiar discussion.
The aim of this post is to ask for a more exhaustive presentation of the insight given by the most-voted answer.
I loved the idea depicted by the author with a story, about finiteness encapsulating the notions of compactness and discreteness, but I was a little bit disappointed about it ending without a demonstration/examples of this fact.
Being someone who learns by categorizing, I was enthusiastic about the idea of thinking about finiteness as unifying the concepts of compactness and discreteness, but I remained with this intuition without being able to realize why this was true or made sense.
So my hope is that through this post I can permit someone to delve deeper on this topic. What are the "compactness attributes" of finiteness? What are the "discreteness attributes" of finiteness?
general-topology analysis compactness
general-topology analysis compactness
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
edited Jan 30 at 10:15
Gabriele Scarlatti
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
asked Jan 29 at 16:50
Gabriele ScarlattiGabriele Scarlatti
380212
380212
2
$begingroup$
I recommend proving many examples are not compact to see what happens when you can’t find any finite subcover that works in order to sharpen intuition about the concept. E.g. Try to show $mathbb{N}$ and $(0,1]$ are not compact in $mathbb{R}$, though maybe you’ve done this already and are still grasping for the idea. Just a suggestion.
$endgroup$
– LoveTooNap29
Jan 29 at 16:55
$begingroup$
I read this article which talks about it, I'll post it maybe it will be useful to someone: blogs.scientificamerican.com/roots-of-unity/…
$endgroup$
– Gabriele Scarlatti
Jan 29 at 16:57
add a comment |
2
$begingroup$
I recommend proving many examples are not compact to see what happens when you can’t find any finite subcover that works in order to sharpen intuition about the concept. E.g. Try to show $mathbb{N}$ and $(0,1]$ are not compact in $mathbb{R}$, though maybe you’ve done this already and are still grasping for the idea. Just a suggestion.
$endgroup$
– LoveTooNap29
Jan 29 at 16:55
$begingroup$
I read this article which talks about it, I'll post it maybe it will be useful to someone: blogs.scientificamerican.com/roots-of-unity/…
$endgroup$
– Gabriele Scarlatti
Jan 29 at 16:57
2
2
$begingroup$
I recommend proving many examples are not compact to see what happens when you can’t find any finite subcover that works in order to sharpen intuition about the concept. E.g. Try to show $mathbb{N}$ and $(0,1]$ are not compact in $mathbb{R}$, though maybe you’ve done this already and are still grasping for the idea. Just a suggestion.
$endgroup$
– LoveTooNap29
Jan 29 at 16:55
$begingroup$
I recommend proving many examples are not compact to see what happens when you can’t find any finite subcover that works in order to sharpen intuition about the concept. E.g. Try to show $mathbb{N}$ and $(0,1]$ are not compact in $mathbb{R}$, though maybe you’ve done this already and are still grasping for the idea. Just a suggestion.
$endgroup$
– LoveTooNap29
Jan 29 at 16:55
$begingroup$
I read this article which talks about it, I'll post it maybe it will be useful to someone: blogs.scientificamerican.com/roots-of-unity/…
$endgroup$
– Gabriele Scarlatti
Jan 29 at 16:57
$begingroup$
I read this article which talks about it, I'll post it maybe it will be useful to someone: blogs.scientificamerican.com/roots-of-unity/…
$endgroup$
– Gabriele Scarlatti
Jan 29 at 16:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're misreading the answer there. It's not that finiteness + discreteness = compactness, but rather that finiteness = discreteness + compactness:
It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.
(emph. mine). Keep in mind that there are lots of topological spaces which are compact but very far from discrete, such as the closed interval $[0,1]$.
So unfortunately, this question is misdirected. Rather, I think the right things to ask are:
What are some ways in which compactness doesn't include discreteness? We can approach this by considering specific compact non-discrete spaces, like $[0,1]$, but a better approach is to realize that any topology coarser than a compact topology is compact. So the further away from discreteness we get, the more likely we are to be compact, in a precise sense!
How does compactness generalize finiteness? This is ultimately more subtle. I think a good starting point here is the bounded value property: a space $mathcal{X}$ has the BVP if any continuous function from $mathcal{X}$ to $mathbb{R}$ has bounded image. (I'm writing this instead of the extreme value theorem because I don't want to bring the completeness of $mathbb{R}$ into it.) Every finite space has BVP trivially - just count the values and compare - but BVP for $[0,1]$ amounts to showing that $[0,1]$ is compact.
Here's how to prove that $[0,1]$ has the BVP. Supposing $f:[0,1]rightarrowmathbb{R}$ is continuous, for $rinmathbb{R}$ let $$U_r={xin [0,1]: f(x)<r}.$$ Clearly (from the continuity of $f$) the family ${U_r:rinmathbb{R}}$ is an open cover of $[0,1]$, so it has a finite subcover, and this in turn implies that we have $[0,1]subseteq U_r$ for some single $r$ (think about how $U_r$ and $U_s$ are related for $r,s$ reals ...).
And here's a brief intuition for how that generalizes the finite situation: we imagine ourselves "counting up" $mathbb{R}$, finding more and more values of $f$ larger than any value seen so far. However, compactness essentially tells us that we can't keep counting forever - in some sense, we "run out" of new values in the same way that we run out of new values when going through a genuinely finite set.
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
$endgroup$
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
add a comment |
$begingroup$
Before we discuss compactness, let's see a noncompact set. Consider $(0,1)$. It's easy to see that
$$(0,1)=bigcup_{n=2}^{infty}(0,1-frac{1}{n})$$
Obviously, removing any $minmathbb{N}$ means that we're going to miss all real numbers $x$ such that $1-frac{1}{m}leq x < 1$. So, it's impossible to find a finite subcover and the definition of compactness fails.
Now let's see what things can go wrong when we're dealing with a noncompact set. Consider the function $log: (0,1) to (-infty,0)$. Even though it's continuous, it is not uniformly continuous because as we get closer to $0$, the slope gets steeper and steeper.
Another issue is that $(0,1)$ is not complete. Meaning that we can find a Cauchy sequence in $(0,1)$ that converges to a point that is not in the space itself, for example the sequence ${frac{1}{n}}_{n=1}^{infty}$. Another issue that can happen is that the image of a noncompact set can become very very big. For example, consider $tan(frac{pi}{2}(2x-1))$ on $(0,1)$ and see that its image has no maximum or minimum value.
Now here is why compact sets are interesting in analysis:
In many proofs, dealing with a compact set means that no matter how we cover it, we can always find a finite subcover. Then we can restrict our attention to the issues that happen in each covering set in the subcover and glue our analysis together to cover the whole set and arrive at a complete proof. This is one of the many ways where finiteness plays a big role. A famous proof that uses this idea is the proof that the Lebesgue measure of $[0,1]$ is equal to $1$.
Compact sets in general have very desirable properties. I will name only a few here:
- A continuous function on a compact set is uniformly continuous.
- Compact sets are very close to being complete. In fact, a metric space is sequentially compact if and only if it is complete and totally bounded.
- In compact sets, every sequence has a convergent subsequence. (This is sometimes taken to be the definition because of Lebesgue's lemma)
- The image of a compact set is compact and hence, it is bounded. So, it has minimum/maximum.
- A continuous invertible function with a compact domain has a continuous inverse.
Edit: as pointed out in the comments, note that these properties are obviously true for finite sets.
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
$endgroup$
2
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
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2 Answers
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2 Answers
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$begingroup$
You're misreading the answer there. It's not that finiteness + discreteness = compactness, but rather that finiteness = discreteness + compactness:
It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.
(emph. mine). Keep in mind that there are lots of topological spaces which are compact but very far from discrete, such as the closed interval $[0,1]$.
So unfortunately, this question is misdirected. Rather, I think the right things to ask are:
What are some ways in which compactness doesn't include discreteness? We can approach this by considering specific compact non-discrete spaces, like $[0,1]$, but a better approach is to realize that any topology coarser than a compact topology is compact. So the further away from discreteness we get, the more likely we are to be compact, in a precise sense!
How does compactness generalize finiteness? This is ultimately more subtle. I think a good starting point here is the bounded value property: a space $mathcal{X}$ has the BVP if any continuous function from $mathcal{X}$ to $mathbb{R}$ has bounded image. (I'm writing this instead of the extreme value theorem because I don't want to bring the completeness of $mathbb{R}$ into it.) Every finite space has BVP trivially - just count the values and compare - but BVP for $[0,1]$ amounts to showing that $[0,1]$ is compact.
Here's how to prove that $[0,1]$ has the BVP. Supposing $f:[0,1]rightarrowmathbb{R}$ is continuous, for $rinmathbb{R}$ let $$U_r={xin [0,1]: f(x)<r}.$$ Clearly (from the continuity of $f$) the family ${U_r:rinmathbb{R}}$ is an open cover of $[0,1]$, so it has a finite subcover, and this in turn implies that we have $[0,1]subseteq U_r$ for some single $r$ (think about how $U_r$ and $U_s$ are related for $r,s$ reals ...).
And here's a brief intuition for how that generalizes the finite situation: we imagine ourselves "counting up" $mathbb{R}$, finding more and more values of $f$ larger than any value seen so far. However, compactness essentially tells us that we can't keep counting forever - in some sense, we "run out" of new values in the same way that we run out of new values when going through a genuinely finite set.
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
$endgroup$
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
add a comment |
$begingroup$
You're misreading the answer there. It's not that finiteness + discreteness = compactness, but rather that finiteness = discreteness + compactness:
It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.
(emph. mine). Keep in mind that there are lots of topological spaces which are compact but very far from discrete, such as the closed interval $[0,1]$.
So unfortunately, this question is misdirected. Rather, I think the right things to ask are:
What are some ways in which compactness doesn't include discreteness? We can approach this by considering specific compact non-discrete spaces, like $[0,1]$, but a better approach is to realize that any topology coarser than a compact topology is compact. So the further away from discreteness we get, the more likely we are to be compact, in a precise sense!
How does compactness generalize finiteness? This is ultimately more subtle. I think a good starting point here is the bounded value property: a space $mathcal{X}$ has the BVP if any continuous function from $mathcal{X}$ to $mathbb{R}$ has bounded image. (I'm writing this instead of the extreme value theorem because I don't want to bring the completeness of $mathbb{R}$ into it.) Every finite space has BVP trivially - just count the values and compare - but BVP for $[0,1]$ amounts to showing that $[0,1]$ is compact.
Here's how to prove that $[0,1]$ has the BVP. Supposing $f:[0,1]rightarrowmathbb{R}$ is continuous, for $rinmathbb{R}$ let $$U_r={xin [0,1]: f(x)<r}.$$ Clearly (from the continuity of $f$) the family ${U_r:rinmathbb{R}}$ is an open cover of $[0,1]$, so it has a finite subcover, and this in turn implies that we have $[0,1]subseteq U_r$ for some single $r$ (think about how $U_r$ and $U_s$ are related for $r,s$ reals ...).
And here's a brief intuition for how that generalizes the finite situation: we imagine ourselves "counting up" $mathbb{R}$, finding more and more values of $f$ larger than any value seen so far. However, compactness essentially tells us that we can't keep counting forever - in some sense, we "run out" of new values in the same way that we run out of new values when going through a genuinely finite set.
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
$endgroup$
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
add a comment |
$begingroup$
You're misreading the answer there. It's not that finiteness + discreteness = compactness, but rather that finiteness = discreteness + compactness:
It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.
(emph. mine). Keep in mind that there are lots of topological spaces which are compact but very far from discrete, such as the closed interval $[0,1]$.
So unfortunately, this question is misdirected. Rather, I think the right things to ask are:
What are some ways in which compactness doesn't include discreteness? We can approach this by considering specific compact non-discrete spaces, like $[0,1]$, but a better approach is to realize that any topology coarser than a compact topology is compact. So the further away from discreteness we get, the more likely we are to be compact, in a precise sense!
How does compactness generalize finiteness? This is ultimately more subtle. I think a good starting point here is the bounded value property: a space $mathcal{X}$ has the BVP if any continuous function from $mathcal{X}$ to $mathbb{R}$ has bounded image. (I'm writing this instead of the extreme value theorem because I don't want to bring the completeness of $mathbb{R}$ into it.) Every finite space has BVP trivially - just count the values and compare - but BVP for $[0,1]$ amounts to showing that $[0,1]$ is compact.
Here's how to prove that $[0,1]$ has the BVP. Supposing $f:[0,1]rightarrowmathbb{R}$ is continuous, for $rinmathbb{R}$ let $$U_r={xin [0,1]: f(x)<r}.$$ Clearly (from the continuity of $f$) the family ${U_r:rinmathbb{R}}$ is an open cover of $[0,1]$, so it has a finite subcover, and this in turn implies that we have $[0,1]subseteq U_r$ for some single $r$ (think about how $U_r$ and $U_s$ are related for $r,s$ reals ...).
And here's a brief intuition for how that generalizes the finite situation: we imagine ourselves "counting up" $mathbb{R}$, finding more and more values of $f$ larger than any value seen so far. However, compactness essentially tells us that we can't keep counting forever - in some sense, we "run out" of new values in the same way that we run out of new values when going through a genuinely finite set.
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
$endgroup$
You're misreading the answer there. It's not that finiteness + discreteness = compactness, but rather that finiteness = discreteness + compactness:
It turns out that finiteness, which you think of as one concept (in the same way that you think of "Foo" as one concept above), is really two concepts: discreteness and compactness. You've never seen these concepts separated before, though. When people say that compactness is like finiteness, they mean that compactness captures part of what it means to be finite in the same way that shortness captures part of what it means to be Foo.
(emph. mine). Keep in mind that there are lots of topological spaces which are compact but very far from discrete, such as the closed interval $[0,1]$.
So unfortunately, this question is misdirected. Rather, I think the right things to ask are:
What are some ways in which compactness doesn't include discreteness? We can approach this by considering specific compact non-discrete spaces, like $[0,1]$, but a better approach is to realize that any topology coarser than a compact topology is compact. So the further away from discreteness we get, the more likely we are to be compact, in a precise sense!
How does compactness generalize finiteness? This is ultimately more subtle. I think a good starting point here is the bounded value property: a space $mathcal{X}$ has the BVP if any continuous function from $mathcal{X}$ to $mathbb{R}$ has bounded image. (I'm writing this instead of the extreme value theorem because I don't want to bring the completeness of $mathbb{R}$ into it.) Every finite space has BVP trivially - just count the values and compare - but BVP for $[0,1]$ amounts to showing that $[0,1]$ is compact.
Here's how to prove that $[0,1]$ has the BVP. Supposing $f:[0,1]rightarrowmathbb{R}$ is continuous, for $rinmathbb{R}$ let $$U_r={xin [0,1]: f(x)<r}.$$ Clearly (from the continuity of $f$) the family ${U_r:rinmathbb{R}}$ is an open cover of $[0,1]$, so it has a finite subcover, and this in turn implies that we have $[0,1]subseteq U_r$ for some single $r$ (think about how $U_r$ and $U_s$ are related for $r,s$ reals ...).
And here's a brief intuition for how that generalizes the finite situation: we imagine ourselves "counting up" $mathbb{R}$, finding more and more values of $f$ larger than any value seen so far. However, compactness essentially tells us that we can't keep counting forever - in some sense, we "run out" of new values in the same way that we run out of new values when going through a genuinely finite set.
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
edited Jan 29 at 18:53
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
answered Jan 29 at 18:46
Noah SchweberNoah Schweber
128k10151294
128k10151294
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
add a comment |
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
$begingroup$
I edited the title to mean what I really meant. I was probably very tired and made a shameful confusion :( Thanks for pointing it out and answering to the right question.
$endgroup$
– Gabriele Scarlatti
Jan 30 at 10:16
add a comment |
$begingroup$
Before we discuss compactness, let's see a noncompact set. Consider $(0,1)$. It's easy to see that
$$(0,1)=bigcup_{n=2}^{infty}(0,1-frac{1}{n})$$
Obviously, removing any $minmathbb{N}$ means that we're going to miss all real numbers $x$ such that $1-frac{1}{m}leq x < 1$. So, it's impossible to find a finite subcover and the definition of compactness fails.
Now let's see what things can go wrong when we're dealing with a noncompact set. Consider the function $log: (0,1) to (-infty,0)$. Even though it's continuous, it is not uniformly continuous because as we get closer to $0$, the slope gets steeper and steeper.
Another issue is that $(0,1)$ is not complete. Meaning that we can find a Cauchy sequence in $(0,1)$ that converges to a point that is not in the space itself, for example the sequence ${frac{1}{n}}_{n=1}^{infty}$. Another issue that can happen is that the image of a noncompact set can become very very big. For example, consider $tan(frac{pi}{2}(2x-1))$ on $(0,1)$ and see that its image has no maximum or minimum value.
Now here is why compact sets are interesting in analysis:
In many proofs, dealing with a compact set means that no matter how we cover it, we can always find a finite subcover. Then we can restrict our attention to the issues that happen in each covering set in the subcover and glue our analysis together to cover the whole set and arrive at a complete proof. This is one of the many ways where finiteness plays a big role. A famous proof that uses this idea is the proof that the Lebesgue measure of $[0,1]$ is equal to $1$.
Compact sets in general have very desirable properties. I will name only a few here:
- A continuous function on a compact set is uniformly continuous.
- Compact sets are very close to being complete. In fact, a metric space is sequentially compact if and only if it is complete and totally bounded.
- In compact sets, every sequence has a convergent subsequence. (This is sometimes taken to be the definition because of Lebesgue's lemma)
- The image of a compact set is compact and hence, it is bounded. So, it has minimum/maximum.
- A continuous invertible function with a compact domain has a continuous inverse.
Edit: as pointed out in the comments, note that these properties are obviously true for finite sets.
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
$endgroup$
2
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
add a comment |
$begingroup$
Before we discuss compactness, let's see a noncompact set. Consider $(0,1)$. It's easy to see that
$$(0,1)=bigcup_{n=2}^{infty}(0,1-frac{1}{n})$$
Obviously, removing any $minmathbb{N}$ means that we're going to miss all real numbers $x$ such that $1-frac{1}{m}leq x < 1$. So, it's impossible to find a finite subcover and the definition of compactness fails.
Now let's see what things can go wrong when we're dealing with a noncompact set. Consider the function $log: (0,1) to (-infty,0)$. Even though it's continuous, it is not uniformly continuous because as we get closer to $0$, the slope gets steeper and steeper.
Another issue is that $(0,1)$ is not complete. Meaning that we can find a Cauchy sequence in $(0,1)$ that converges to a point that is not in the space itself, for example the sequence ${frac{1}{n}}_{n=1}^{infty}$. Another issue that can happen is that the image of a noncompact set can become very very big. For example, consider $tan(frac{pi}{2}(2x-1))$ on $(0,1)$ and see that its image has no maximum or minimum value.
Now here is why compact sets are interesting in analysis:
In many proofs, dealing with a compact set means that no matter how we cover it, we can always find a finite subcover. Then we can restrict our attention to the issues that happen in each covering set in the subcover and glue our analysis together to cover the whole set and arrive at a complete proof. This is one of the many ways where finiteness plays a big role. A famous proof that uses this idea is the proof that the Lebesgue measure of $[0,1]$ is equal to $1$.
Compact sets in general have very desirable properties. I will name only a few here:
- A continuous function on a compact set is uniformly continuous.
- Compact sets are very close to being complete. In fact, a metric space is sequentially compact if and only if it is complete and totally bounded.
- In compact sets, every sequence has a convergent subsequence. (This is sometimes taken to be the definition because of Lebesgue's lemma)
- The image of a compact set is compact and hence, it is bounded. So, it has minimum/maximum.
- A continuous invertible function with a compact domain has a continuous inverse.
Edit: as pointed out in the comments, note that these properties are obviously true for finite sets.
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
$endgroup$
2
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
add a comment |
$begingroup$
Before we discuss compactness, let's see a noncompact set. Consider $(0,1)$. It's easy to see that
$$(0,1)=bigcup_{n=2}^{infty}(0,1-frac{1}{n})$$
Obviously, removing any $minmathbb{N}$ means that we're going to miss all real numbers $x$ such that $1-frac{1}{m}leq x < 1$. So, it's impossible to find a finite subcover and the definition of compactness fails.
Now let's see what things can go wrong when we're dealing with a noncompact set. Consider the function $log: (0,1) to (-infty,0)$. Even though it's continuous, it is not uniformly continuous because as we get closer to $0$, the slope gets steeper and steeper.
Another issue is that $(0,1)$ is not complete. Meaning that we can find a Cauchy sequence in $(0,1)$ that converges to a point that is not in the space itself, for example the sequence ${frac{1}{n}}_{n=1}^{infty}$. Another issue that can happen is that the image of a noncompact set can become very very big. For example, consider $tan(frac{pi}{2}(2x-1))$ on $(0,1)$ and see that its image has no maximum or minimum value.
Now here is why compact sets are interesting in analysis:
In many proofs, dealing with a compact set means that no matter how we cover it, we can always find a finite subcover. Then we can restrict our attention to the issues that happen in each covering set in the subcover and glue our analysis together to cover the whole set and arrive at a complete proof. This is one of the many ways where finiteness plays a big role. A famous proof that uses this idea is the proof that the Lebesgue measure of $[0,1]$ is equal to $1$.
Compact sets in general have very desirable properties. I will name only a few here:
- A continuous function on a compact set is uniformly continuous.
- Compact sets are very close to being complete. In fact, a metric space is sequentially compact if and only if it is complete and totally bounded.
- In compact sets, every sequence has a convergent subsequence. (This is sometimes taken to be the definition because of Lebesgue's lemma)
- The image of a compact set is compact and hence, it is bounded. So, it has minimum/maximum.
- A continuous invertible function with a compact domain has a continuous inverse.
Edit: as pointed out in the comments, note that these properties are obviously true for finite sets.
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
$endgroup$
Before we discuss compactness, let's see a noncompact set. Consider $(0,1)$. It's easy to see that
$$(0,1)=bigcup_{n=2}^{infty}(0,1-frac{1}{n})$$
Obviously, removing any $minmathbb{N}$ means that we're going to miss all real numbers $x$ such that $1-frac{1}{m}leq x < 1$. So, it's impossible to find a finite subcover and the definition of compactness fails.
Now let's see what things can go wrong when we're dealing with a noncompact set. Consider the function $log: (0,1) to (-infty,0)$. Even though it's continuous, it is not uniformly continuous because as we get closer to $0$, the slope gets steeper and steeper.
Another issue is that $(0,1)$ is not complete. Meaning that we can find a Cauchy sequence in $(0,1)$ that converges to a point that is not in the space itself, for example the sequence ${frac{1}{n}}_{n=1}^{infty}$. Another issue that can happen is that the image of a noncompact set can become very very big. For example, consider $tan(frac{pi}{2}(2x-1))$ on $(0,1)$ and see that its image has no maximum or minimum value.
Now here is why compact sets are interesting in analysis:
In many proofs, dealing with a compact set means that no matter how we cover it, we can always find a finite subcover. Then we can restrict our attention to the issues that happen in each covering set in the subcover and glue our analysis together to cover the whole set and arrive at a complete proof. This is one of the many ways where finiteness plays a big role. A famous proof that uses this idea is the proof that the Lebesgue measure of $[0,1]$ is equal to $1$.
Compact sets in general have very desirable properties. I will name only a few here:
- A continuous function on a compact set is uniformly continuous.
- Compact sets are very close to being complete. In fact, a metric space is sequentially compact if and only if it is complete and totally bounded.
- In compact sets, every sequence has a convergent subsequence. (This is sometimes taken to be the definition because of Lebesgue's lemma)
- The image of a compact set is compact and hence, it is bounded. So, it has minimum/maximum.
- A continuous invertible function with a compact domain has a continuous inverse.
Edit: as pointed out in the comments, note that these properties are obviously true for finite sets.
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
edited Jan 30 at 20:06
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
answered Jan 29 at 18:31
stressed outstressed out
6,5531939
6,5531939
2
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
add a comment |
2
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
2
2
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
$begingroup$
Note to the OP: note that all these properties are trivially true for finite subsets. Compactness is a way to generalise properties of finite sets to more general sets.
$endgroup$
– Henno Brandsma
Jan 29 at 23:00
add a comment |
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I recommend proving many examples are not compact to see what happens when you can’t find any finite subcover that works in order to sharpen intuition about the concept. E.g. Try to show $mathbb{N}$ and $(0,1]$ are not compact in $mathbb{R}$, though maybe you’ve done this already and are still grasping for the idea. Just a suggestion.
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– LoveTooNap29
Jan 29 at 16:55
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I read this article which talks about it, I'll post it maybe it will be useful to someone: blogs.scientificamerican.com/roots-of-unity/…
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– Gabriele Scarlatti
Jan 29 at 16:57