MHB Open Sets - Unions and Intersections -

[Math Amateur]

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

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:View attachment 7181In 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 ...

Peter*** EDIT ***

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 prove this intuition is correct ...

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

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The above exercise relies on the definition of open sets and related concepts and so to provide readers with a knowledge of Sohrab's definitions and notation I am provided the following text ...https://www.physicsforums.com/attachments/7182

 

[Evgeny.Makarov]

After some reflection I am beginning to believe that $$\bigcap_{ n = 1}^{ \infty } = [0,1]$$ where I$$_n = [ \frac{1}{n}, 1 - \frac{1}{n} ]$$

Do you mean $$\bigcap_{ n = 1}^{ \infty }I_n=[0,1]$$? This is not correct. The intersection of any family $\mathcal{F}$ of sets is a subset of any set $S\in\mathcal{F}$. That's the meaning of intersection: it contains only those element that belong to all sets in the family. Now, $I_2=\left[\frac12,1-\frac12\right]=\left\{\frac12\right\}$; therefore, $\bigcap_{n=1}^\infty I_n\subseteq \{1/2\}$.

" ... ... 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 ... ... "

Why don't you use the suggested family of intervals instead of $I_n$ above?

I am unable to make a meaningful start on this problem ... can someone help me with the exercise

As I said, $\bigcap\mathcal{F}\overset{\text{def}}{=}\{x\mid \forall S\in F\,x\in S\}$. Find $\bigcap_{n=1}^\infty\left(-\frac{1}{n} , 1 + \frac{1}{n}\right)$ by definition. Consider the cases $x<0$, $0\le x\le 1$ and $x>1$ and find if $x$ belongs to every interval in the family.

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

This problem does not involve limits or convergence. It involves the definition of union and intersection of arbitrary families of sets.

 

[Math Amateur]

Do you mean $$\bigcap_{ n = 1}^{ \infty }I_n=[0,1]$$? This is not correct. The intersection of any family $\mathcal{F}$ of sets is a subset of any set $S\in\mathcal{F}$. That's the meaning of intersection: it contains only those element that belong to all sets in the family. Now, $I_2=\left[\frac12,1-\frac12\right]=\left\{\frac12\right\}$; therefore, $\bigcap_{n=1}^\infty I_n\subseteq \{1/2\}$.

Why don't you use the suggested family of intervals instead of $I_n$ above?

As I said, $\bigcap\mathcal{F}\overset{\text{def}}{=}\{x\mid \forall S\in F\,x\in S\}$. Find $\bigcap_{n=1}^\infty\left(-\frac{1}{n} , 1 + \frac{1}{n}\right)$ by definition. Consider the cases $x<0$, $0\le x\le 1$ and $x>1$ and find if $x$ belongs to every interval in the family.

This problem does not involve limits or convergence. It involves the definition of union and intersection of arbitrary families of sets.

Hi Evgeny ...

My post contained a serious typo ... caused by careless cutting and pasting ...

My ***EDIT*** should have read ... as follows:" ... *** EDIT ***

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

Note that Sohrab doesn't define limits or convergence until after setting this exercise ..."I still believe $$\bigcap_{ n = 1}^{ \infty } I_n = [0,1]$$ is correct ... but still cannot prove it ..Thanks for your hints and guidance ... but I need further help ...

Can you help further ...

Peter

 

[Math Amateur]

Hi Evgeny ...

My post contained a serious typo ... caused by careless cutting and pasting ...

My ***EDIT*** should have read ... as follows:" ... *** EDIT ***

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 prove this intuition is correct ...

Note that Sohrab doesn't define limits or convergence until after setting this exercise ..."I still believe $$\bigcap_{ n = 1}^{ \infty } I_n = [0,1]$$ is correct ... but still cannot prove it ..Thanks for your hints and guidance ... but I need further help ...

Can you help further ...

Peter

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:https://www.physicsforums.com/attachments/7188

 

[Math Amateur]

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:

Hmm ... just been pointed out to me (on the Physics Forums) that the intersection being closed doesn't mean it is not open ...

Oh well ... back to the drawing board ...

Peter

 

[Evgeny.Makarov]

Your proof that $\bigcap_{n=1}^\infty I_n=[0,1]$ is correct. The interval $[0,1]$ is not open because $0$ does not have a neighborhood that is contained entirely in the interval. If you prove that $[0,1]$ is closed, it is also sufficient because Example 2.2.3 says that $\emptyset$ and $\mathbb{R}$ are the only sets that are open and closed at the same time (this needs to be proved, though).

 

[Math Amateur]

Your proof that $\bigcap_{n=1}^\infty I_n=[0,1]$ is correct. The interval $[0,1]$ is not open because $0$ does not have a neighborhood that is contained entirely in the interval. If you prove that $[0,1]$ is closed, it is also sufficient because Example 2.2.3 says that $\emptyset$ and $\mathbb{R}$ are the only sets that are open and closed at the same time (this needs to be proved, though).

Thanks Evgeny ... indeed! see your point!

Nevertheless ...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, 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? Feel free to critique proof if necessary ...

Peter

 

[Evgeny.Makarov]

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?)

This is a step backwards. First you should prove all reasonable properties of real numbers (such as the axioms of an ordered field and their corollaries) and then proceed to the study of calculus or topology. Proving the inequality you are asking about requires something like $\forall x,y>0\,\forall z<0\,x<y\implies yz<xz$. The proof of this statement depends on how real numbers were introduced (axiomatically or constructively).

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

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

You probably mean $x<-1/m$.

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

$$-x \gt \frac{1}{m}$$ follows from $x<-1/m$ (my version) and not $$x \gt - \frac{1}{m}$$ (your version). Also, $$-x \gt \frac{1}{m}$$ follows from $x<-1/m$ by arithmetic and not by Archimedean property. You probably want to use the Archimedean property to justify the existence of $m$ such that $x<-1/m$. The proof must proceed from what is given, and each step should be justified. If you say, "I claim $P$ which would follow from $Q$, and I am going to use $R$ to justify $Q$", this creates a spaghetti-like structure in the proof. Instead, it should be: "We have $R$; therefore, $Q$, from which follows $P$".

Thus for $$n \gt m$$ ... that is $$- \frac{1}{n} \gt - \frac{1}{m}$$

I would add: $\ldots>x$ to make it clear that $x$ indeed lies outside of $$\left( - \frac{1}{n}, 1 + \frac{1}{n}\right)$$.

Also it would probably be better to consider some fixed $x$ such as $-\epsilon/2$, but this is not really important.

 

[Math Amateur]

This is a step backwards. First you should prove all reasonable properties of real numbers (such as the axioms of an ordered field and their corollaries) and then proceed to the study of calculus or topology. Proving the inequality you are asking about requires something like $\forall x,y>0\,\forall z<0\,x<y\implies yz<xz$. The proof of this statement depends on how real numbers were introduced (axiomatically or constructively).

You probably mean $x<-1/m$.

$$-x \gt \frac{1}{m}$$ follows from $x<-1/m$ (my version) and not $$x \gt - \frac{1}{m}$$ (your version). Also, $$-x \gt \frac{1}{m}$$ follows from $x<-1/m$ by arithmetic and not by Archimedean property. You probably want to use the Archimedean property to justify the existence of $m$ such that $x<-1/m$. The proof must proceed from what is given, and each step should be justified. If you say, "I claim $P$ which would follow from $Q$, and I am going to use $R$ to justify $Q$", this creates a spaghetti-like structure in the proof. Instead, it should be: "We have $R$; therefore, $Q$, from which follows $P$".

I would add: $\ldots>x$ to make it clear that $x$ indeed lies outside of $$\left( - \frac{1}{n}, 1 + \frac{1}{n}\right)$$.

Also it would probably be better to consider some fixed $x$ such as $-\epsilon/2$, but this is not really important.

Thanks for the critique ... and sorry for a couple of typos ...

Really appreciate your feedback ... it is really helpful ...

Peter