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I am working on a proof of in which I have to assume infinity. Am I allowed to assume the reader is familiar with how limits work?

Cro-Magnon
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    In my opinion, your proof has gaps: for example how do you know $\bigcup_{n=1}^{\infty}[3-\frac{1}{n}, 6] - \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] =[2,3)$. One may expect this to be proven as well. Also you start with $x$, but no where in your proof the $x$ appears until the very end, where you all of a sudden claim $x \in[3,6]$. – Anurag A Sep 15 '18 at 21:55
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    Why do you claim: $$\bigcup_{n=1}^{\infty}[3-\frac{1}{n}, 6] - \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] =[2,3)$$ instantly? Proving this statement would be of similar difficulty as to proving the statement you're after. Hint: apply the direct definition of $x$ being an element of the intersection. – Stan Tendijck Sep 15 '18 at 21:58
  • @stanTendijck I will include it, one moment – Cro-Magnon Sep 15 '18 at 22:02
  • @stanTenDijck, I included it as lemma. Not entirely correct, but should be enough to show how I proved that. I am working on a problem where I have to prove multiple things. – Cro-Magnon Sep 15 '18 at 22:06
  • @AnuragA I have included it as a lemma. – Cro-Magnon Sep 15 '18 at 22:09
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    There are still a lot of flaws in that reasoning. Let me point it out in an answer below – Stan Tendijck Sep 15 '18 at 22:21
  • @StanTendijck Thank you! – Cro-Magnon Sep 15 '18 at 22:24
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    To the 2nd trial: in the 1st part, you've assumed $x$ something or another, but you never used $x$ in the process, so they could be discarded. The 2nd part was still not rigorous. Why $[3,6]$ is "the only interval we can be sure is in every $A_m$"? Why cannot it be $[4,6]$? This might not be some big deal to the conclusion, but this assertion is not supported. – xbh Sep 16 '18 at 12:52
  • @xbh Thank you. I will first try to get the second part then return for contrapositive of the first. For the second part I'm just unsure why $\forall m \geq 1, 3-\frac{1}{m} < 3$ isn't enough to say exclude [4, 6]. I feel like I understand what I have to show but that I don't have enough vocabulary to show $lim_{m \rightarrow \infty} 3- \frac{1}{m}=3$ and therefore only [3, 6] is possible – Cro-Magnon Sep 16 '18 at 13:34
  • @xbh How about; Given that the intersection only contains elements shared by all and as we approach infinity $3- \frac{1}{m}$ gets close to 3 but never equal to 3 it follows [3,6]? – Cro-Magnon Sep 16 '18 at 13:40
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    @Cro-Magnon Demo: since $x \in \cap A_n$, $x \in [3-1/n, 6]$ for all $n$, then $6 \geqslant x \geqslant 3-1/n$ for all $n$, let $n \to \infty$ we get $6 \geqslant x \geqslant 3$. // Here i never write that $[3,6]$ is the only interval. – xbh Sep 16 '18 at 13:42
  • @xbh I changed my answer there from 'is the only interval in every $A_{m}$' to 'is shared by every $A_{m}$'. I think this is what I meant and is similar to what you have in your demo (I think). Because $m \in \mathbb{N}$ and we stated $m \geq 1$ which implies $m \rightarrow \infty$ . My apologies for the confusion, I've been working since yesterday on these problems :P Can't let them go – Cro-Magnon Sep 16 '18 at 13:49
  • @xbh Ah, I see the issue. They're different. It's simple but I complicate it. I will continue working on it. Thanks for your time. Hopefully these posts will help some other people also. – Cro-Magnon Sep 16 '18 at 13:59
  • @Cro-Magnon You are welcome. If you want to save this for future reference, then I suggest you please take some time to create an answer and post all these discussions after some summarization, since the comment region is generally not for this. Otherwise these might be deleted in future due to various reasons. – xbh Sep 16 '18 at 14:03
  • @xbh Mind if I ask you to look one time at the answer I provided? I followed your guidelines as much as possible. If this one is wrong I will continue making exercises but will leave this one for the future when I'm more confident. – Cro-Magnon Sep 16 '18 at 17:04
  • The third line of your post needs a re-write. When you say "So,..." it suggests ypu are stating a conclusion. What you mean to say is, "So to prove..." – DanielWainfleet Sep 16 '18 at 22:58
  • @DanielWainfleet Could you look at my answer at the bottom ? (Not in my question but in the answers.) I think it is complete and correct. https://math.stackexchange.com/a/2919138/238010 – Cro-Magnon Sep 17 '18 at 07:54
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    @Cro-Magnon .It is correct. In the 2nd line you do not need to say "where $n\in \Bbb N$" because $\cap_{n=1}^{\infty}$ always means $\cap_{n\in \Bbb N}.$ I assume the student may assume that $(\forall n \in \Bbb N;(3-1/n\leq x))\iff 3\leq x$ and not be required to prove it. – DanielWainfleet Sep 17 '18 at 20:57

2 Answers2

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In this answer, I would like to focus on your Lemma (since proving this is similar to your case and since I don't like providing direct solutions, you still gotta work for it). Let me also emphasize that if you have proven this lemma, you still are not done by far. If I were you, I would try to prove the statement similar as to the lemma which I will show below.

So, you want to prove that $\bigcup_{n=1}^{\infty} [3-n^{-1},6] = [2,6]$.

Let $x\in\bigcup_{n=1}^{\infty} [3-n^{-1},6]$. Then you claim something which is not very well formulated (since you are not using standard mathematical reasoning), I would do it as follows. Hence, per definition, there exists an $m\in\{1,2,\dots\}$ such that $x\in[3-m^{-1},6]$. Since $m\geq 1$, we have $3-m^{-1}\geq 2$ and thus $[3-m^{-1},6]\subseteq [2,6]$ which implies that $x\in[2,6]$. This is approximately similar to your reasoning, however, this is way more direct and using the definitions.

The other way around is treated wrongly by you. You would have to argue as follows. Let $x\in[2,6] = [3-1^{-1},6]$ and thus $x\in \bigcup_{n=1}^{\infty} [3 - n^{-1},6]$ since this set would only be bigger.

Stan Tendijck
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    One extra notion for proving your statement. In proving that $$\bigcap_{n=1}^{\infty} [3-1/n,6]\subseteq[3,6],$$ you need to worry a little bit about the boundary. – Stan Tendijck Sep 15 '18 at 22:34
  • Thank you that really helped me! I will keep working on it till I get it right. – Cro-Magnon Sep 15 '18 at 22:37
  • Could you take a look at my partial try in my question? I just want to know whether I am on right path. It's hard to figure out what definitions to use and how much argumentation to give. – Cro-Magnon Sep 15 '18 at 23:25
  • I think I finally understand your answer. I removed the partial because it was wrong. Thanks a lot! – Cro-Magnon Sep 16 '18 at 00:02
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    You're welcome. Have a nice day! – Stan Tendijck Sep 16 '18 at 07:17
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We have to prove the following.

$ \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [3, 6] $ where $ n \in \mathbb{N}$

This means $x \in \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \iff x \in [3,6]$

Proof;

Let $x \in \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6]$

Then $3- \frac{1}{n} \leq x \leq 6$ for all n. Now, let $n \rightarrow \infty$ then it follows , $3 \leq x \leq 6$

Hence $x \in [3,6]$ and $ \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \subseteq [3,6]$

For the reverse conditional we have,

let $x \in [3,6]$ then $6 \geq x \geq 3$.

Given that $3-\frac{1}{n} < 3$ for all n, it follows that $x \geq 3 > 3 - \frac{1}{n}$

and hence $[3 - \frac{1}{n}, 6] $ is a bigger interval than [3, 6] for all n.

Therefore [3, 6] $ \subseteq \bigcap_{n=1}^{\infty}A_{n}$ and $x \in \bigcap_{n=1}^{\infty}A_{n}$

$\therefore \bigcap_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [3, 6] \blacksquare$

Cro-Magnon
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