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Following is Topology Without Tears by Morris

Let A,B be a space of R with Euclidean topology If A is the set of all rational numbers,B set of irrational number

Consider these 4 sets :

(1)A $\cap$ cl(B) (2)cl(A)$ \cap$ B. (3)cl(A)$\cap$ cl(B)(4 ) cl(A$\cap$B)

Prove that no two sets are equal

(This is last best attempt. Please upvote me. If it is wrong, show the proof. The proof in full has not been given to date .When I tag it I put #proofexplanation or something similar. This implies I want explanation. I am having a hell of time learning it. I might not be crying, but am close to bashing my head against the wall in agony

In my previous attempt I thought I was dealing with dense sets since Q and R/Q are dense in R so that is how I did it..)

Try

0.Q and R$\setminus$ Q are the same under closure in R

1.Since A $\cap B =\emptyset$ ,cl(A $\cap B)=\emptyset$ Furthermore A and B are dense in R.

2.(From :Interior, closure and boundary of the sets of all rational and irrational numbers.)

Whenever A & B are dense disjoint subsets of a topological space X we have cl(A)=X=cl(B) by Def.of being dense. Since B$\subset A^{c}$ and A$\subset B^{c}$ it follows cl($A^{c})=X=cl(B^{c}$)

Since we have A being the set of rationals (Q) ,B being the set of irrationals (R\Q) and set intersection and taking compliments we get:

(i) A$\cap$cl(B) = Q $\cap$ R=Q

(ii)cl(A)$\cap $B= R $\cap$ R\Q= R\Q

(iii)= R $\cap$ R=R

(iv) ={}

Therefore none of them are equal

  • I think you have to tell us what $Q$ means and what $I$ means. – Gerry Myerson Jan 15 '21 at 02:07
  • OK. Please edit that into the body of the question. People shouldn't need to read the comments to understand what the question is asking. And what does it mean when you put a bar over the union symbol? Do you mean $\overline{\overline A\cup\overline B}$? – Gerry Myerson Jan 15 '21 at 02:30
  • Also, you have $A,B$ as subsets of $R$, but as elements of $Q,I$. That doesn't make sense. – Gerry Myerson Jan 15 '21 at 02:34
  • Actually, the whole thing is nonsense. $X$ is never defined. It is never given that $B$ is dense in $X$. Please start over, being more careful. – Gerry Myerson Jan 15 '21 at 02:35
  • I rewrote the theorem to be proved. I am not to good with these proofs.@GerryMyerson –  Jan 15 '21 at 02:40
  • Isn't $(1)$ and (2) the same? both are $\mathbb{R}$, no? Check – AyamGorengPedes Jan 15 '21 at 02:48
  • I don't understand what union bar is, so I won't comment for $(4)$, but I think $(1),(2),(3) = \mathbb{R}$. But $(1),(2),(3)$ would be different if it's intersection instead of union. – AyamGorengPedes Jan 15 '21 at 02:51
  • Yup, then check ing with (3) and (4) should prove it –  Jan 15 '21 at 02:51
  • "dense" is an adjective, not an operator. Please clarify. – Trevor Gunn Jan 15 '21 at 02:55
  • In topology and related areas of mathematics, a subset A of a topological space X is called dense (in X) if every point x in X either belongs to A or is a limit point of A; that is, the closure of A constitutes the whole set X From Wikipedia –  Jan 15 '21 at 02:59
  • I know what the adjective "dense" means. What does the operator $\bar{A}$ mean? The closure? The set of limit points? The complement? – Trevor Gunn Jan 15 '21 at 03:00
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    @larrymintz If the question is correct (that no 2 out of the 4 union are equal), then $(1) \neq (2) \neq (3) \neq (4)$. Since as you have written, $(1) = (2)$, I believe there may have been a misunderstanding on what the question is actually asking. Can you type exactly, word by word, the question? – AyamGorengPedes Jan 15 '21 at 03:14
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    The current statement is still incomprehensible. It's not clear whether $\bar A$ means the complement of $A$ or the closure of $A$. It's not clear what it means when you put a bar over the intersection symbol. I don't know what "according to community wiki" means. I don't know why you write $B\subset A$ when that's clearly false. You're still introducing the symbol $X$ without telling us what it means. Mathematics is about communication. – Gerry Myerson Jan 15 '21 at 06:08
  • I got rid of the bar symbol and put cl() instead,so now the question is clear. ,I hope. I rewrote my whole proof,given the hints.I know math is about communication. When I mentioned the “community wiki” I thought you could go to it and see what I was referring to. Morris uses URL’s to,so maybe he is not legit,< giggles >@GerryMyerson –  Jan 15 '21 at 15:41
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    OK, it looks much better. I still don't know what you mean by "community wiki", and I don't know how I could "go to it". I didn't see any URLs. "Since $A\cap B$ is empty, ${\rm cl\ }A\cap{\rm\ cl\ }B$ is empty" is false. But mostly what you've written looks good. – Gerry Myerson Jan 15 '21 at 21:20
  • I can’t remember “community wiki “ now But I what I stated came from here https://math.stackexchange.com/questions/283987/interior-closure-and-boundary-of-the-sets-of-all-rational-and-irrational-number . Can I be upvoted. I spent a ton of time on it –  Jan 15 '21 at 22:46
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    I suspect that the users who downvoted have moved on to other things and don't know about the improvements, and other users are looking at questions that have been posted more recently and have pushed yours off the front pages. Perhaps you could be happy that you got a good answer and maybe learned a few things on the way. – Gerry Myerson Jan 17 '21 at 00:18

1 Answers1

2

You copied the problem in the book badly. (Particularly, you flipped $\cap$ as $\cup$, which makes the conclusion completely wrong.)

Here is the original one (one page 78, Chapter 3):

enter image description here

and the bar means "closure" (Definition 3.1.9 on page 75).

You are asking 6(a):

  • The set in (i) is equal to $A$
  • The set in (ii) is equal to $B$
  • The set in (iii) is equal to $\mathbb{R}$
  • The set in (iv) is empty.

So no two of the above four sets are equal.

Notes.

You would be able to understand the answer if you read the relevant section of the book carefully.

  • Example 3.1.12 tells you that $\overline{\mathbb{Q}}=\mathbb{R}$. This is also mentioned in Section 3.4.
  • It is a good exercise for you to work out that the set of irrational numbers is also dense in $\mathbb{R}$.
  • It does not make sense to write $\bar{A}\bar{\cap}\bar{B}$. A correct way to type (iv) in the problem is $\overline{A\cap B}$.
  • It does? I know I am asking 6(a).your right .Me bad. –  Jan 15 '21 at 03:27
  • I never realized the fact that the closure of the empty set is the empty set. It's 4.30am and I'm giggling. – AyamGorengPedes Jan 15 '21 at 03:29
  • But you don’t explain how and that is what I want to know –  Jan 15 '21 at 04:42
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    The way you posted the question, larry, just asked for the answers. Now that you have the answers, now you tell us that's not enough – you really want the explanations. Your question needs to be rewritten from top to bottom. – Gerry Myerson Jan 15 '21 at 06:11
  • Yes I read Example 3.1.12. As to third point,I didn’t know the code for it I just knew \bar As second, I will do it for R\Q. Sometimes I find it hard to see how all the theorems work together. In real/complex analysis I saw it .I am self-studying this field.. –  Jan 15 '21 at 16:11
  • The whole post has been rewritten. I can’t get it any better or explain myself any better @GerryMyerson –  Jan 15 '21 at 16:45