Open Sets - Unions and Intersections - Sohrab Ex. 2.2.4 (1)

[Math Amateur]

Homework Statement

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with a part of Exercise 2.2.4 Part (1) ... ...

Exercise 2.2.4 Part (1) reads as follows:

In the above text from Sohrab we read the following:

" ... ... Using the infinite collection $( \frac{ -1 }{n} , 1 + \frac{ 1 }{n} ), \ n \in \mathbb{N}$, show the latter statement is false if $\Lambda$ is infinite ... ... "I am unable to make a meaningful start on this problem ... can someone help me with the exercise ...

Homework Equations

Sohrab's definition of \epsilon neighborhoods and open and closed sets are relevant ... so I am providing these as follows:

The Attempt at a Solution

[/B]
After some reflection I am beginning to believe that $\bigcap_{ n = 1}^{ \infty } I_n = [0,1]$ where $I_n = ( \frac{-1}{n}, 1 + \frac{1}{n} )$ ... but ... sadly ... I cannot (rigorously) prove this intuition is correct ...

Note that Sohrab doesn't define limits or convergence until after setting this exercise ...

 

[fresh_42]

For a point $x \in \bigcup_{\lambda \in \Lambda}\mathcal{O}_\lambda$ you need to find an open ball $B_\varepsilon(x)$ completely in this set.
Now what does it mean for $x$ to be in this union and what do you know about a single $\mathcal{O}_\lambda \,?$

The same line of proof can show that $\bigcap_{\lambda \in \Lambda}\mathcal{O}_\lambda$ with $|\Lambda| < \infty$ is open. Find an open ball around some arbitrary point of the set, which is entirely within it.

For the last part, better look at $I_n = (1-\frac{1}{n},1+\frac{1}{n})$ and build the intersection of all.

If you want to take the example given, it is clear that $[0,1] \subseteq \bigcap_{n \in \mathbb{N}}(-\frac{1}{n},1+\frac{1}{n})$. Now we need equality. So assume a point $x \notin [0,1]$. Can it be in all sets $(-\frac{1}{n},1+\frac{1}{n})\,?$

 

[andrewkirk]

A standard way to prove that a union of sets has some property that each set in the union has is to observe that, by definition, if $x\in \bigcup_{\lambda\in\Lambda} O_\lambda$ then there exists at least one set $O_a$, where $a\in\Lambda$ such that $x\in O_a$.
Look at the definition of an open set and see if you can use what it tells you about the relationship between $x$ and $O_a$ to say something about the relationship between $x$ and $\bigcup_{\lambda\in\Lambda} )_\lambda$.

Similarly, to prove something about an intersection of sets, use the fact that if $x\in \bigcap_{\lambda\in\Lambda} O_\lambda$ then $x$ is in every one of the $O_\lambda$. Again use the definition of open set to see what that tells you about the relationship between $x$ and each of the $O_\lambda$. Can you use that to say something similar about the relationship between $x$ and the intersection? The fact that the intersection is finite will be critical, as we will be dealing with a bunch of Open Balls of different sizes, centred on $x$ and we want to find an open ball that fits inside them all.

 

[Math Amateur]

Hi fresh_42, Andrew ...

My apologies but I seem to have misled you ... the only part of Exercise 2.2.4 (1) that I have a problem with is the one stating:

" ... ... Using the infinite collection $( \frac{ -1 }{n} , 1 + \frac{ 1 }{n} ), \ n \in \mathbb{N}$, show the latter statement is false if $\Lambda$ is infinite ... ... "

Can you say more about how to solve this part of the exercise ...

Peter

 

[fresh_42]

See above!

It is clear that $[0,1] \subseteq \bigcap_{n \in \mathbb{N}}(-\frac{1}{n},1+\frac{1}{n})$. Now we need equality. So assume a point $x \notin [0,1]$. Can it be in all sets $(-\frac{1}{n},1+\frac{1}{n})\,?$ Construct a set $(-\frac{1}{k},1+\frac{1}{k})$ that doesn't contain $x$.
This will show what you already suspected, that the intersection equals $[0,1]\,.$

Now what's left is to show, that $[0,1]$ is closed, i.e. $(-\infty,0) \cup (1,\infty)$ is open. Choose a point in this union and find an open ball which is still within it.

 

[Math Amateur]

See above!

It is clear that $[0,1] \subseteq \bigcap_{n \in \mathbb{N}}(-\frac{1}{n},1+\frac{1}{n})$. Now we need equality. So assume a point $x \notin [0,1]$. Can it be in all sets $(-\frac{1}{n},1+\frac{1}{n})\,?$ Construct a set $(-\frac{1}{k},1+\frac{1}{k})$ that doesn't contain $x$.
This will show what you already suspected, that the intersection equals $[0,1]\,.$

Now what's left is to show, that $[0,1]$ is closed, i.e. $(-\infty,0) \cup (1,\infty)$ is open. Choose a point in this union and find an open ball which is still within it.

Try to show that $\bigcap_{ n = 1}^{ \infty } I_n = [0,1]$

where $I_n = ( \frac{-1}{n}, 1 + \frac{1}{n} )$ ... ...We need to show that $[0,1] \subseteq \bigcap_{ n = 1}^{ \infty } I_n$ and $\bigcap_{ n = 1}^{ \infty } I_n \subseteq [0,1]$ ...

Clearly $[0,1] \subseteq \bigcap_{ n = 1}^{ \infty } I_n$ so what we need to show rigorously is $\bigcap_{ n = 1}^{ \infty } I_n \subseteq [0,1]$ ...To show $\bigcap_{ n = 1}^{ \infty } I_n \subseteq [0,1]$ ... we need to show

$x \in \bigcap_{ n = 1}^{ \infty } I_n \Longrightarrow x \in [0,1]$ ... ... ... ... (1)Show contrapositive of (1) ...

... visually ... $x \notin [0,1] \Longrightarrow x \notin \bigcap_{ n = 1}^{ \infty } I_n$

and consider the two case: $x \lt 0$ and $x \gt 1$ ... ...$x\lt 0$

If $x \lt 0$ then $\exists m$ such that $x \lt \frac{ -1 }{m}$ (Corollary 2.1.31 (see below) to Archimedean Property of $\mathbb{R}$ since we are requiring $-x \gt \frac{1}{m}$ where $-x \gt 0$ ...)

Thus $x \notin ( \frac{-1}{n}, 1 + \frac{1}{n} )$ for $n \ge m$

... and since there are sets (for some $n \in \mathbb{N})$ which do not contain $x$, the intersection $\bigcap_{ n = 1}^{ \infty } I_n$ does not contain $x$ ... ...
$x \gt 1$

If $x \gt 1$ then $\exists m$ such that $x \gt 1 + \frac{1}{m}$ (Corollary 2.1.31 (see below) to Archimedean Property of $\mathbb{R}$ since we are requiring $(x - 1 ) \gt \frac{1}{m}$ and we have $(x - 1) \gt 0$ ... )

Thus $x \notin ( \frac{-1}{n}, 1 + \frac{1}{n} )$ for $n \ge m$

... and since there are sets (for some $n \in \mathbb{N}$ which do not contain $x$, the intersection $\bigcap_{ n = 1}^{ \infty } I_n$ does not contain $x$ ... ...
Thus for all $x \notin [0, 1]$ we have $x \notin \bigcap_{ n = 1}^{ \infty } I_n$ ... ... which is what we wanted to show ...So $\bigcap_{ n = 1}^{ \infty } I_n = [0,1]$ which is closed ...Is that correct?

Peter
Notes: (1) Strictly, we need to prove $[0,1]$ is closed ... but that seems straightforward ...

(2) The above post mentions the Archimedean Property and it Corollary (Corollary 2.1.31) ... so I am providing a scan of these as follows:

 

[andrewkirk]

@Math Amateur
Sorry to be a nit picker but perhaps you'll forgive me since this nit pick leads to a shorter proof.

What the author has asked us to show is that an infinite intersection of open sets is not necessarily open.
In topology, being closed is neither a necessary nor a sufficient condition for not being open. In the discrete topology, all sets are both open and closed. Even in metric topologies like this, the empty and universal sets are both open and closed.

Fortunately, proving a set is not open is - at least in this case - easier than proving it is closed. All we need to do is find a point that is not an 'interior point', ie which cannot be the centre of any open ball that is contained in the set. Intuitively we expect both 0 and 1 to be such points, since they are boundaries of the interval that we suspect equals the infinite intersection.

To prove the theorem doesn't work for infinite intersections, all we need to do is:

1. Show that 0 lies inside every set that is being intersected (easy), from which it follows, by definition, that it lies in the intersection.

2. Show that, for any $\epsilon>0$, there is some $m$ such that the ball $B_\epsilon(0)$ lies at least partly outside $I_m$. It should be easy to express such a $m$ in terms of $\epsilon$.

We don't need to prove that the infinite intersection is the interval $[0,1]$. All we need is to suspect that is the case, which gives us the idea of investigating open balls around the boundary point 0.

 

[Math Amateur]

@Math Amateur
Sorry to be a nit picker but perhaps you'll forgive me since this nit pick leads to a shorter proof.

What the author has asked us to show is that an infinite intersection of open sets is not necessarily open.
In topology, being closed is neither a necessary nor a sufficient condition for not being open. In the discrete topology, all sets are both open and closed. Even in metric topologies like this, the empty and universal sets are both open and closed.

Fortunately, proving a set is not open is - at least in this case - easier than proving it is closed. All we need to do is find a point that is not an 'interior point', ie which cannot be the centre of any open ball that is contained in the set. Intuitively we expect both 0 and 1 to be such points, since they are boundaries of the interval that we suspect equals the infinite intersection.

To prove the theorem doesn't work for infinite intersections, all we need to do is:

1. Show that 0 lies inside every set that is being intersected (easy), from which it follows, by definition, that it lies in the intersection.

2. Show that, for any $\epsilon>0$, there is some $m$ such that the ball $B_\epsilon(0)$ lies at least partly outside $I_m$. It should be easy to express such a $m$ in terms of $\epsilon$.

We don't need to prove that the infinite intersection is the interval $[0,1]$. All we need is to suspect that is the case, which gives us the idea of investigating open balls around the boundary point 0.

Hi Andrew ...

Oh! Didn't stop to think carefully ... you're right of course ...

Have to get the energy to start again ...

Thanks for pointing that out Andrew ...

Peter

 

[Math Amateur]

@Math Amateur
Sorry to be a nit picker but perhaps you'll forgive me since this nit pick leads to a shorter proof.

What the author has asked us to show is that an infinite intersection of open sets is not necessarily open.
In topology, being closed is neither a necessary nor a sufficient condition for not being open. In the discrete topology, all sets are both open and closed. Even in metric topologies like this, the empty and universal sets are both open and closed.

Fortunately, proving a set is not open is - at least in this case - easier than proving it is closed. All we need to do is find a point that is not an 'interior point', ie which cannot be the centre of any open ball that is contained in the set. Intuitively we expect both 0 and 1 to be such points, since they are boundaries of the interval that we suspect equals the infinite intersection.

To prove the theorem doesn't work for infinite intersections, all we need to do is:

1. Show that 0 lies inside every set that is being intersected (easy), from which it follows, by definition, that it lies in the intersection.

2. Show that, for any $\epsilon>0$, there is some $m$ such that the ball $B_\epsilon(0)$ lies at least partly outside $I_m$. It should be easy to express such a $m$ in terms of $\epsilon$.

We don't need to prove that the infinite intersection is the interval $[0,1]$. All we need is to suspect that is the case, which gives us the idea of investigating open balls around the boundary point 0.

Thanks again, Andrew ...

...Will try a direct proof that $\bigcap_{n=1}^\infty I_n$ is not open ...

Step (1): Show that $0$ lies inside every interval being intersected ...

For $n \gt 0$ where $n \in \mathbb{N}$, we have that $\frac{1}{n} \gt 0$ ... that is ... $0 \gt - \frac{1}{n}$ (but! how do I justify this?)

Therefore $0 \in \bigcap_{n=1}^\infty I_n$

[Not sure if the above is valid and justified ... wonder if there is a better argument?]Step (2): Show that for every $\epsilon \gt 0, \epsilon \in \mathbb{R}, \exists m$ such that $B_{ \epsilon } (0)$ lies partly outside of $I_n$ for $n \gt m$ ... ...

Let $\epsilon \gt 0$ where $\epsilon \in \mathbb{R}$ ...

Consider $x \in B_{ \epsilon } (0)$ such that $- \epsilon \lt x \lt 0$

Now ... $\exists m$ such that $x \gt - \frac{1}{m}$

That is $-x \gt \frac{1}{m}$ (Archimedean Property 0 Corollary 2.1.32 (b) )Thus for $n \gt m$ ... that is $- \frac{1}{n} \gt - \frac{1}{m}$ we have that $x$ lies outside of the interval/set $( - \frac{1}{n}, 1 + - \frac{1}{n} )$ ...

... and hence $B_{ \epsilon } (0)$ is not contained within $\bigcap_{n=1}^\infty I_n$ ...Since this holds for arbitrary $\epsilon \gt 0$, we have that $\bigcap_{n=1}^\infty I_n$ is not open ...
Is that correct? Please feel free to critique proof ...

Peter

 

[andrewkirk]

It looks pretty good.

Step (1): Show that $0$ lies inside every interval being intersected ...

For $n \gt 0$ where $n \in \mathbb{N}$, we have that $\frac{1}{n} \gt 0$ ... that is ... $0 \gt - \frac{1}{n}$ (but! how do I justify this?)

What is needed here is an axiom or a theorem that says
$$x<y\to x+a<y+a\ \ (P)$$

What axioms and theorems does the text give us about inequalities, that may enable us to derive $P$ as a theorem?

 

[Math Amateur]

It looks pretty good.What is needed here is an axiom or a theorem that says
$$x<y\to x+a<y+a\ \ (P)$$

What axioms and theorems does the text give us about inequalities, that may enable us to derive $P$ as a theorem?

Hi Andrew ...

You write:

" ... ... What axioms and theorems does the text give us about inequalities, that may enable us to derive $P$ as a theorem? ... ... "Well ... Sohrab defines inequalities in the following text ... which includes an exercise requesting students to prove that:

$a \lt b \Longrightarrow a + c \lt b + c$

Sohrab also has the following Theorem and Corollary concerning inequalities:

Hope that helps,

Peter

 

[andrewkirk]

Well ... Sohrab defines inequalities in the following text ... which includes an exercise requesting students to prove that:

$a \lt b \Longrightarrow a + c \lt b + c$

Great, that's exactly what's needed to justify that step you weren't sure about:

For $n \gt 0$ where $n \in \mathbb{N}$, we have that $\frac{1}{n} \gt 0$ ... that is ... $0 \gt - \frac{1}{n}$ (but! how do I justify this?)

Assuming that the step you want to justify is
$$\frac{1}{n} \gt 0\ \ \to\ \ 0 \gt - \frac{1}{n}$$
we just substitute in the above theorem $\frac1n$ for $b$, $0$ for $a$ and $-\frac1n$ for $c$.

however, you can't just jump straight to the conclusion $0 \in \bigcap_{n=1}^\infty I_n$ of your Step 1, because you've only shown that 0 is above the lower bound of $I_n$. You need to do the same sort of thing to show that 0.

In Step 2, there's just a bit of a muddle over signs:

Step (2):

Consider $x \in B_{ \epsilon } (0)$ such that $- \epsilon \lt x \lt 0$

Now ... $\exists m$ such that $x \gt - \frac{1}{m}$

That is $-x \gt \frac{1}{m}$ (Archimedean Property 0 Corollary 2.1.32 (b) )

The 'that is' multiplies both sides of the inequality by -1, which means we have to reverse the direction of the inequality.

I find dealing with lots of negative signs confusing, because of that sort of thing inter alia, and hence prefer to leave negation as late as possible in a proof.

I suggest that instead you choose $x\in (0,\epsilon)$, find $m\in\mathbb N$ such that $x>\frac1m$, then show that $-x$ lies inside the ball $B_\epsilon(0)$ and outside $I_m$.