Limit points, closure of set (Is my proof correct?)

[Incand]

Homework Statement

Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $E$ and $\bar E = E \cup E'$ have the same limit points. Do $E$ and $E'$ always have the same limit points?

Homework Equations

Theorem:
(i) $\bar E$ is closed
(ii) $E=\bar E$ if and only $E$ is closed.

Theorem:
For any collection of $\{G_\alpha\}$ of open sets, $\bigcup_\alpha G_\alpha$ is open.

The Attempt at a Solution

I tried to prove the above but since my solution seems quite different from the one in the solution manual I suspect I may have done something wrong. I Would appreciate if someone could take a look and see if it's wrong/incomplete or if I done something unnecessarily complicated.

We split the first part of proof into two cases:
Case 1: Assume $E$ is open. Then $E'$ has to be closed since the union of two open sets is open and $\bar E$ is closed (contradiction otherwise).

Case 2: Assume $E$ is closed. Then $E= \bar E$. Then it follows that $(\bar E)' = E'$ but then $(E')' \subseteq E'$. Hence every limit point of $E'$ is in $E'$ so $E'$ is closed.Take any $x\in E'$ then $x \in \bar E$. Since $x$ is a limit point of $E$ and $E \subseteq \bar E$, $x$ is also a limit point of $\bar E$. We then have $E' \subseteq (\bar E)'$.
Take $x \in (\bar E)'$. Since $(\bar E)'$ is closed (as we showed before) this means $x \in \bar E$. Then at least one of the following is true $x \in E$ or $x \in E'$. If the second case is true we're done.

Otherwise if $x \in E$. We can further assume $E$ is open (If $E$ is closed we're done since then $E = \bar E$). This means that there is an open ball around $x$ $B(x,r) = \{y;d(x,y)<r\}$ that is a subset of $E$ and hence $\bar E$. Then since $x$ is a limit point of $\bar x$ that open ball must contain another point $q \in E$ such that $q \ne p$.

But then every open ball with a larger radius must also contain this point. If instead the radius is smaller that open ball is contained in $B$ and hence in $\bar E$ and hence that ball must also contain a point $q_2 \in E$ such that $q_2 \ne x$. Hence $x \in E'$ and $E' = (\bar E)'$.

 

[fresh_42]

Without having read your solution in detail: closed and open are not negations of each other: ]0,1] is neither.
If A is closed (open) you may only conclude that its complement is open (closed), but not open isn't closed.
In addition I don't see that you are allowed to use a metric.

 

[Incand]

Without having read your solution in detail: closed and open are not negations of each other: ]0,1] is neither.
If A is closed (open) you may only conclude that its complement is open (closed)

Thanks for taking a look at it! I was aware of that fact but I guess the words tricked me. Seems I have to at least find another proof of the first part.
As for using a metric, our book only works in metric spaces so I think I'm allowed to assume them.

 

[fresh_42]

Perhaps you could start with a formal definition of what it means for a set to be closed, and how a limit point is defined.
It's always useful to simply clarify the conditions and assumptions first. Often it will lead you automatically to the right idea.

 

[Incand]

Alright. A set if closed if every limit point of the set is in the set. A point $p$ is a limit point if for every open ball around $x$, that is $B_M(p,r) = \{y; d(p,y)<r\}$ contains a point $q \in E$ such that $q \ne p$.

We want to show that $E'$ is closed. That means that every limit point of $E'$ is in $E'$, i.e. $(E')' \subseteq E'$.

So I made another attempt using the above:
Take $x\in (E')'$ then there is a $q\in E'$ such that $0<d(x,q)<r_1$. Since $q \in E'$ we also have $0<d(q,y)<r_2$ for some $y\in E$. We want to show that $\forall r>0$, $0<d(x,y)<r$.
From the triangle inequality
$d(x,y) \le d(x,q)+d(y,q) < r_1 +r_2 = r$.
But we also have that
$d(x,q) \le d(x,y)+d(q,y) \Longrightarrow d(x,y) \ge d(x,q)-d(q,y) > d(x,y)-r_2$. Take $r_2 = d(x,y)$ then $d(x,y)> 0$. Combined this means that $x\in E'$.

 

[fresh_42]

Alright. A set if closed if every limit point of the set is in the set. A point $p$ is a limit point for every if every open ball around $x$, that is $B_M(p,r) = \{y; d(p,y)<r\}$ contains a point $q \in E$ such that $q \ne p$.

Ok, that provides us a common base on how the terms are used (for there are different ways to define them).

We want to show that $E'$ is closed. That means that every limit point of $E'$ is in $E'$, i.e. $(E')' \subseteq E'$.

So I made another attempt using the above:
Take $x\in (E')'$ then

for any $r_1>0$

there is a $q\in E'$ such that $0<d(x,q)<r_1$. Since $q \in E'$ we also have

for any $r_2>0$

$0<d(q,y)<r_2$ for some $y\in E$.

Personally I think it would be better to enumerate $q=q_{r_1}$ and $y=y_{r_2}$ because they depend on your choice of the radii $r_i$.

We want to show that $\forall r>0$, $0<d(x,y)<r$.

This cannot be shown for $x$ and $y$ are fixed and $r$ is not.
You want to show that for any $r>0$ there is a $y_r \in E$ with $0<d(x,y_r)<r$.

From the triangle inequality
$d(x,y) \le d(x,q)+d(y,q) < r_1 +r_2 = r$.

We now know that $r>0$ is given. Arbitrary small chosen, but given. How should you chose $r_1$ and $r_2$ then?

I assume the rest now is to show that $x \neq y$ to rule out the constant sequence.

But we also have that
$d(x,q) \le d(x,y)+d(q,y) \Longrightarrow d(x,y) \ge d(x,q)-d(q,y) > d(x,y)-r_2$. Take $r_2 = d(x,y)$ then $d(x,y)> 0$. Combined this means that $x\in E'$.

I think you cheated here. $x$ and $y$ are found to given radii $r_i$. If you fix $r_2$ how can you still get arbitrary close?
And you didn't use the facts $x \neq q$ and $y \neq q$. What would it mean if $x \in E$?

 

[fresh_42]

You want to show that $x \in E'$.
What does it mean if this were not the case? Can you then still find arbitrary close points $y \in E$ like you've shown above?

 

[Incand]

You want to show that $x \in E'$.
What does it mean if this were not the case? Can you then still find arbitrary close points $y \in E$ like you've shown above?

No, if there was an arbitrary close point $y\in E$ then $y$ would be a limit point by definition since it's in every open ball around $x$.Restating the first part of the proof with your corrections: (Btw. isn't choosing $r_1,r_2$ so it works out nicely mostly for aesthetics?)
Take $x\in (E')'$. Then $\forall r>0\; \exists q_r \in E', q_r \ne x$ such that $0<d(x,q_r)<r/2$.
Since $q_r \in E'$ we have that $\forall r>0 \; \exists y_r \in E, y_r\ne q_r$ such that $0<d(q_r,y_r)<r/2$.

From the triangle inequality
$d(x,y_r) \le d(x,q_r)+d(y_r,q_r) < r/2 +r/2 = r$ for all $r >0$. So there is always some $q_r$ within the open ball around $x$.

I assume the rest now is to show that $x \neq y$ to rule out the constant sequence.

I think you cheated here. $x$ and $y$ are found to given radii $r_i$. If you fix $r_2$ how can you still get arbitrary close?
And you didn't use the facts $x \neq q$ and $y \neq q$. What would it mean if $x \in E$?

Agreed. I'm trying to think this through. I'm not sure I understand your last question. It would mean $(E')' \subseteq E$, that is the limit points of $E'$ is in $E$?

 

[fresh_42]

No, if there was an arbitrary close point $y\in E$ then $y$ would be a limit point by definition since it's in every open ball around $x$.Restating the first part of the proof with your corrections: (Btw. isn't choosing $r_1,r_2$ so it works out nicely mostly for aesthetics?)
Take $x\in (E')'$. Then $\forall r>0\; \exists q_r \in E', q_r \ne x$ such that $0<d(x,q_r)<r/2$.
Since $q_r \in E'$ we have that $\forall r>0 \; \exists y_r \in E, y_r\ne q_r$ such that $0<d(q_r,y_r)<r/2$.

From the triangle inequality
$d(x,y_r) \le d(x,q_r)+d(y_r,q_r) < r/2 +r/2 = r$ for all $r >0$. So there is always some $q_r$ within the open ball around $x$.

I agree. One $r$ looks better. I simply didn't want to change your choice.

Agreed. I'm trying to think this through. I'm not sure I understand your last question. It would mean $(E')' \subseteq E$, that is the limit points of $E'$ is in $E$?

For $x$ to be a limit point (of $E$) we have to find arbitrary close points of $E$ which aren't $x$ itself. The detour over $q$ doesn't guarantee us that the chosen $y$ isn't always the $x$ we started from. Since $y \in E$ we are done if $x \notin E$. So what happens to an $x \in E$? If we assume $x\notin E'$ (in order to achieve a contradiction) we would get $x$ as an isolated point of $E$, i.e. there would be an open ball $B_M(x,r_0)$ with no element of $E$ but $x$. But why can't this happen? Or may it happen?
Edit: Correction: Closed ball $\overline{B_M(x,r_o)}$ but this doesn't make any difference because one can always find a little smaller open one within. Btw: your last $q_r$ should be a $y_r$.

 

[Ray Vickson]

Homework Statement

Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $E$ and $\bar E = E \cup E'$ have the same limit points. Do $E$ and $E'$ always have the same limit points?

Homework Equations

Theorem:
(i) $\bar E$ is closed
(ii) $E=\bar E$ if and only $E$ is closed.

Theorem:
For any collection of $\{G_\alpha\}$ of open sets, $\bigcup_\alpha G_\alpha$ is open.

The Attempt at a Solution

I tried to prove the above but since my solution seems quite different from the one in the solution manual I suspect I may have done something wrong. I Would appreciate if someone could take a look and see if it's wrong/incomplete or if I done something unnecessarily complicated.

We split the first part of proof into two cases:
Case 1: Assume $E$ is open. Then $E'$ has to be closed since the union of two open sets is open and $\bar E$ is closed (contradiction otherwise).

Case 2: Assume $E$ is closed. Then $E= \bar E$. Then it follows that $(\bar E)' = E'$ but then $(E')' \subseteq E'$. Hence every limit point of $E'$ is in $E'$ so $E'$ is closed.Take any $x\in E'$ then $x \in \bar E$. Since $x$ is a limit point of $E$ and $E \subseteq \bar E$, $x$ is also a limit point of $\bar E$. We then have $E' \subseteq (\bar E)'$.
Take $x \in (\bar E)'$. Since $(\bar E)'$ is closed (as we showed before) this means $x \in \bar E$. Then at least one of the following is true $x \in E$ or $x \in E'$. If the second case is true we're done.

Otherwise if $x \in E$. We can further assume $E$ is open (If $E$ is closed we're done since then $E = \bar E$). This means that there is an open ball around $x$ $B(x,r) = \{y;d(x,y)<r\}$ that is a subset of $E$ and hence $\bar E$. Then since $x$ is a limit point of $\bar x$ that open ball must contain another point $q \in E$ such that $q \ne p$.

But then every open ball with a larger radius must also contain this point. If instead the radius is smaller that open ball is contained in $B$ and hence in $\bar E$ and hence that ball must also contain a point $q_2 \in E$ such that $q_2 \ne x$. Hence $x \in E'$ and $E' = (\bar E)'$.

Your arguments assume that you are dealing with sets in a metric space. What happens if you are in a general topological space, where there is not necessarily any metric $d(x,y)$?

If restriction to a metric space is "understood" from course context, for example, then OK, you do not need to say more. However, if you are not assuming metric spaces by default, then you should at least explain that your proof is, perhaps, valid only in a metric space.

 

[Incand]

I agree. One $r$ looks better. I simply didn't want to change your choice.

For $x$ to be a limit point (of $E$) we have to find arbitrary close points of $E$ which aren't $x$ itself. The detour over $q$ doesn't guarantee us that the chosen $y$ isn't always the $x$ we started from. Since $y \in E$ we are done if $x \notin E$. So what happens to an $x \in E$? If we assume $x\notin E'$ (in order to achieve a contradiction) we would get $x$ as an isolated point of $E$, i.e. there would be an open ball $B_M(x,r_0)$ with no element of $E$ but $x$. But why can't this happen? Or may it happen?
Edit: Correction: Closed ball $\overline{B_M(x,r_o)}$ but this doesn't make any difference because one can always find a little smaller open one within. Btw: your last $q_r$ should be a $y_r$.

I'm not sure how to prove this either way really or even how to get started. Should I try to show that $x \notin E$ or that $x \in E'$ directly? I obvious want to get to the second one eventually. We know that $x \in (E')'$, I don't know how from that tell if there's an open ball without another element of $E$.

Your arguments assume that you are dealing with sets in a metric space. What happens if you are in a general topological space, where there is not necessarily any metric $d(x,y)$?

If restriction to a metric space is "understood" from course context, for example, then OK, you do not need to say more. However, if you are not assuming metric spaces by default, then you should at least explain that your proof is, perhaps, valid only in a metric space.

Topological spaces are beyond this course and haven't even been mentioned. It's a first course in analysis for first/second year of undergraduate. Perhaps in I a few years time I could take a look at the general case as an exercise :)

 

[fresh_42]

I'm not sure how to prove this either way really or even how to get started. Should I try to show that $x \notin E$ or that $x \in E'$ directly? I obvious want to get to the second one eventually. We know that $x \in (E')'$, I don't know how from that tell if there's an open ball without another element of $E$.

You have already done it. We have to show that $x \in (E´)´$ implies $x \in E'$. You already showed that there are arbitrarily close $y \in E$ to such an $x$. There's only left to show that the $y$ aren't $x$ itself. Since $x \in E'$ is our goal we may assume $x\notin E'$.
But if $x$ is no limit point of $E$ then there is a radius $r_0$ around $x$ with no other point of $E$ in it except perhaps $x$.
Now we can fix an $N \in ℕ$ and a $q_N \in E'$ with $d(q_N,x) < \frac{1}{2} r_0$ and a $y_N \in E$ with $d(y_N,q_N) < \frac{1}{2} r_0$. This special $y_N$, however, is closer than $r_0$ at $x$ contradicting the assumption.
It is the same argument again, only for special points in special distances.

Now you have proven that $E'$ is closed.
Remains to show that $(\overline{E})' = (E ∪ E')' ⊆ E'$ (you already have shown the other direction in the OP) and whether there are examples of $E' ⊂ (E')'$?

Topological spaces are beyond this course and haven't even been mentioned. It's a first course in analysis for first/second year of undergraduate. Perhaps in I a few years time I could take a look at the general case as an exercise :)

In the general case one usually has to assume some kind of a separation axiom, e.g. the topological space to be Hausdorff, i.e. any two points can be separated by disjoint open neighbourhoods. These neighbourhoods then play the role of your balls and the proofs are similar.

 

[Incand]

Now we can fix an $N \in ℕ$ and a $q_N \in E'$ with $d(q_N,x) < \frac{1}{2} r_0$ and a $y_N \in E$ with $d(y_N,q_N) < \frac{1}{2} r_0$. This special $y_N$, however, is closer than $r_0$ at $x$ contradicting the assumption.

I guess you're using the triangle inequality here, that is $d(x,y_N) \le d(x,q_N)+d(q_N,y_N) < r_0/2 + r_2/0 = r_0$ hence closer than $r_0$. I'm wondering why you need to fix an $N\in \mathbb{N}$? Aren't we simply allowed to take a $q$ we want regardless?

 

[Incand]

Now you have proven that $E'$ is closed.
Remains to show that $(\overline{E})' = (E ∪ E')' ⊆ E'$ (you already have shown the other direction in the OP)

I can see two approaches here, either show this directly or show that the limit points of an (finite) union is a subset of the union of the limit points of each element.

Starting with the first approach should be similar to what I've already done.
Take $x \in (E \cup E')'$. Then $\forall r>0 \exists q_r \in \bar E, \; q_r \ne x$ such that $0<d(x,q_r)<r/2$.
If $q_r\in E'$ we exactly as earlier have an $y_r$ in every open ball around $x$ (however perhaps I need to show that $x \ne y_r$ differently).
Else however $q_r \in E$ but then $x \in E'$ since $d(x,q_r)<r/2<r$ ,$x \ne q_r$.

 

[fresh_42]

I guess you're using the triangle inequality here, that is $d(x,y_N) \le d(x,q_N)+d(q_N,y_N) < r_0/2 + r_2/0 = r_0$ hence closer than $r_0$. I'm wondering why you need to fix an $N\in \mathbb{N}$? Aren't we simply allowed to take a $q$ we want regardless?

Sorry, I have been caught in the picture of convergent sequences where choosing $N$ large enough guarantees the existence of the desired $q$. But you are right, it is sufficient to pick a $q$ with $d(x,q) < \frac{r_0}{2}$. My fault. Stuck in the definition I once learned.

 

[fresh_42]

I can see two approaches here, either show this directly or show that the limit points of an (finite) union is a subset of the union of the limit points of each element.

Starting with the first approach should be similar to what I've already done.
Take $x \in (E \cup E')'$. Then $\forall r>0 \exists q_r \in \bar E, \; q_r \ne x$ such that $0<d(x,q_r)<r/2$.
If $q_r\in E'$ we exactly as earlier have an $y_r$ in every open ball around $x$ (however perhaps I need to show that $x \ne y_r$ differently).

No, it's alright because $y_r \in E$ and being different from $x$ is as before.

Else however $q_r \in E$ but then $x \in E'$ since $d(x,q_r)<r/2<r$ ,$x \ne q_r$.

Yes. I don't see anything wrong. It is all about the triangle inequality applied to the special cases.

The last question can be formulated as: Can we think of limit points of $(E')'$ which aren't limit points of $E$ already?
I guess the trick is the same again: If $x$ is a limit point of $E'$, i.e. a limit point of limit points of $E$ then there can be found points in $E$ as well. (Imagine a closed circle. Then the circle itself is the boundary of the open disc. You can have a point on the circle as a limit point of other circle points as well as of points within the disc.)

 

[Incand]

I think you got the last one confused If you meant the last question in the op. We already showed that $(E')' \subseteq E'$. The question is if $(E')'$ may be a proper subset of $E'$. An example of this would be the set $\{x; x=1/n, n \in \mathbb N\}$. The set of limit points is $\{0\}$ so $(E')' = \varnothing$.

There's also examples when $(E')' = E'$. An example would be $\{ x; x\in [0,1], x \in \mathbf R\}$ in $\mathbb R$.

 

[fresh_42]

@Incand Good point!

 

[Incand]

I think we managed to show everything we hoped for! Thanks for all the help, I learned a lot from going through the exercise with your help.