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I think my proof is wrong but I don't know how to approach the statement differently. I hope you can help me identify where I'm mistaken/incomplete.

Proof: $$\text{We need to prove: } \bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [2, 6] $$

$$\text{Thus, } x \in \bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \iff x \in [2, 6]$$

$$\text{We first consider the converse of the biconditional.}$$

$$\text{and proceed by contrapositive.} $$ $$x \notin \bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \implies x \notin [2, 6]$$ $$\text{Given that when } n = 1, [3-\frac{1}{n}, 6]=[2,6] \text{ and } $$ $$ \forall z \in (\mathbb{N} - {1}) , [3-\frac{1}{z}, 6] < [2, 6] \text{ thus } \bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [2,6]$$ $$\text{It follows that, } x \notin [2,6] \text{. Thus the converse is true.}$$

$$\text{Now, for left to right } (\implies) \text{ we proceed by direct proof. }$$

$$x \in \bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] \implies x \in [2, 6]$$ $$\text{By the same logic as for the converse, we continue..}$$

$$\text{Given that, when } n = 1, [3-\frac{1}{n}, 6] = [2, 6], \text{ It follows that: } $$ $$x \in [2,6]$$

$$\therefore \bigcup_{n=1}^{\infty} A_{n} = [2, 6] \text{ } \blacksquare$$

Thank you for your time.

Updated proof:

Proof:

We assume $x \in \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6]$ $$A_{1} = [2, 6] > \bigcup_{i=2}^{\infty} [3 - \frac{1}{i}, 6] = [ \frac{5}{2}, 6] *$$ $$\therefore A_{1} = \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6] = [2, 6] , \space x \in [2,6]$$

[ I placed a (*) to show where I'm uncertain. My problem is in knowing how much I should explain to the reader. I have to establish somehow that $A_{1}$ is the biggest interval but I kind of leave open 'why' $\bigcup_{i=2}^{\infty} [3 - \frac{1}{i}, 6] = [ \frac{5}{2}, 6]$ is true. For example, I thought I had to show why $3 - \frac{1}{i} > 2$ for every i $\geq$ 2. So I have a tedency to break everything down too much]

Now for the converse we proceed by contrapositive.

We assume $x \notin \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6]$ $$A_{1} = [2, 6] > \bigcup_{i=2}^{\infty} [3 - \frac{1}{i}, 6] = [ \frac{5}{2}, 6] *$$ $$\therefore A_{1} = \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6] = [2, 6] , \space x \notin [2,6]$$

$$ \therefore \bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [2, 6] \blacksquare$$

Updated proof #2:

Proof:

We assume, $x \in \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6]$.

Since $2 \leq 3 - \frac{1}{n} < 3$ for all $ n \geq 1$, $ \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6] \subseteq [2, 6], x \in [2, 6]$

For the converse we assume $x \notin \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6]$.

Following the same reasoning as above, Since $2 \leq 3 - \frac{1}{n} < 3$ for all $ n \geq 1$, $ \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6] \subseteq [2, 6], x \notin [2, 6]$

$\therefore \bigcup_{n=1}^{\infty} [3 - \frac{1}{n}, 6] = [2, 6] \space \blacksquare$

Cro-Magnon
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    No the last part with the WLOG is problematic. Given that $x\in \bigcup_n A_n$ doesn't mean that $x\in A_1$. In your situation it works but that's not generally true. – Yanko Sep 15 '18 at 14:36
  • It is definitely wrong to write all of the English parts of the proof inside math mode! – hmakholm left over Monica Sep 15 '18 at 14:40
  • @henningmakholm sorry about that. – Cro-Magnon Sep 15 '18 at 14:41
  • Seems okay, but need to improve. The notations are kind of random [or those are typos, i do not know], and the dots are need to be connected. – xbh Sep 15 '18 at 14:44
  • I've read your proof and given my comments in my answer. Feel free to check it. – xbh Sep 15 '18 at 17:16
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    Seems okay but a little too much. Consider $[2,6] = [3-\frac 1n,6]{n=1}\subset\cup{n=1}^{\infty}[3-\frac 1n,6]$. And for all $n\in\mathbb N$ then $2\le 3-\frac 1n <3<6$ so $[3-\frac 1n, 6]\subset [2,6]$ so $\cup_{n=1}^{\infty}[3-\frac 1n,6]\subset [2,6]$. – fleablood Sep 15 '18 at 17:58
  • @fleablood can you explain why you added < 6 to $2 \leq 3 - \frac{1}{n} \le 3$. Is it also correct to just say $2 \leq 3 - \frac{1}{n} \le 3$? – Cro-Magnon Sep 15 '18 at 19:07
  • @fleablood Would you mind looking at proof update #2? I think it is better than the previous ones. My only question is can I get rid of the redundant "since 2 $\leq$ 3 - \frac{1}{n} < 3" because I have to say it twice. – Cro-Magnon Sep 15 '18 at 19:24
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    I added the $< 6$ just to make it explicitly clear $[3 -\frac 1n,6]$ is a subset $[2,6]$. It wasn't necessary. – fleablood Sep 15 '18 at 22:43
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    In your latest proof I would insert something to the following "Since $2 \le 3-\frac 13 < 3$ ...then $[3-\frac 1n, 6] \subset [2,6]$ for each $n$ and there fore the union of all $[3-\frac n,6]\subset [2,6]$. The converse seems weird to me. Why should $2 \le 3-\frac 1n$ imply that $x \not \in [2,6]$? Suppose $2 \le x < 3- \frac 1n$ for all $n$? Easier to just say $[2,6] = [3-\frac 1n, 6]$ for $n=1$ so $[2,6] \subset $ of any union containing it. – fleablood Sep 15 '18 at 22:52

3 Answers3

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We have $[3-\frac{1}{n},6]\subseteq[2,6]$ for all $n\ge 1$, and thus $$\bigcup_{n=1}^\infty\left[3-\frac{1}{n},6\right]\subseteq[2,6].$$ As for the reverse inclusion, we have $$[2,6]=\left[3-\frac{1}{1},6\right]\subseteq\bigcup_{n=1}^\infty\left[3-\frac{1}{n},6\right].$$

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You should end with something like:$$\text{From }x\in\bigcup_{n=1}^{\infty} \left[3-\frac1n,6\right]\text{ it follows that }x\in\left[3-\frac1n,6\right]\subseteq[2,6]\text{ for some positive integer }n$$

That integer does not have to be $1$.

drhab
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  • My idea was that since, by the well-ordering principle, $\mathbb{N}$ has a least element 1 and for 1 we get the biggest interval, that that should be enough to say that the union is equal to [2,6] and thus x $\notin$ [2,6] – Cro-Magnon Sep 15 '18 at 14:54
  • @Cro-Magnon Yes, but you did not write down them unambiguously. – xbh Sep 15 '18 at 15:15
  • @xbh I will redo the problem and update my question. If you can take a look at it after i've redone it that would be awesome. – Cro-Magnon Sep 15 '18 at 15:18
  • @Cro-Magnon No problem. But add as an update, do not change the whole question. – xbh Sep 15 '18 at 15:21
  • @xbh, ok thanks! – Cro-Magnon Sep 15 '18 at 15:21
  • @xbh I updated the question with a new proof. The (*) refers to where I have difficulties. Somehow that step seems unnecessary because it still implies that the reader knows how the index collection of unions is defined. The text between [ ] is my attempt at explaining this. Thank you for your time. – Cro-Magnon Sep 15 '18 at 16:28
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Not to the question but to the updated proof:

You have not yet proved $$ \bigcup_1^\infty \left[3 -\frac 1n, 6\right] = [2,6], $$ so in your proof such equation is definitely not allowed to appear. Also I still do not clearly get your logic inferences in your proof, i.e. I do not see the reasoning part. Here I write a demonstration.

Demo proof

$\blacktriangleleft$ We show that $$ x \in \bigcup_1^\infty \left[3 - \frac 1n, 6\right] \iff x \in [2,6]. $$

$\implies$ part:

Assume $x \in \bigcup_1^\infty [3-1/n, 6]$, then there exists an $m \in \mathbb N^*$ s.t. $x \in [3-1/m, 6]$ [This is the definition of union]. Since for all $n\in \Bbb N^*$, $[3 -1/n, 6] \subseteq [2,6]$, we get $x \in [2,6]$ as well.

$\impliedby$ part:

Proceed by contrapositive. Suppose $x \notin \bigcup_1^n [3-1/n, 6]$, then $x \notin [3-1/n, 6]$ for all $n \in \Bbb N^*$, then particularly $x \notin [3-1,6] = [2,6]$.

Combined the results we conclude that $$ \bigcup_1^\infty \left[3 - \frac 1n, 6\right] = [2,6]. \blacktriangleright $$

xbh
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  • Thank you! I see where I messed up. I reread the chapter and will continue making problems. – Cro-Magnon Sep 15 '18 at 18:06
  • I added another proof update (#2) I think it is much better. I use different properties than in your example but I think it works well. If it is correct, do you have any idea how I could get rid of the double sentences "since 2 ≤ 3 - \frac{1}{n} < 3 ..." since that seems redundant. – Cro-Magnon Sep 15 '18 at 19:28
  • @Cro-Magnon Logic issue: from $x \in \cup [3-1/n, 6] \implies x \in [2,6]$ we deduce that $\cup [3-1/n, 6]\subseteq [2,6]$, not the reverse direction. In your proof you seems to proceed as the answer of @ eloiPrime, it is unnecessary to assume $x\in \cup [3-1/n,6] $ anymore. The second part does not make sense. $A\subseteq B$ and $x\notin A$ does not imply that $x \notin B$. Again, your reasoning seems "floated", and you make readers to guess what you are thinking. If the question is more complicated, then your proof would be totally vague. – xbh Sep 16 '18 at 03:58
  • Thank you so much. I am starting to understand the importance of using definitions in your proofs. You have helped me a lot. – Cro-Magnon Sep 16 '18 at 08:04
  • @Cro-Magnon You are welcome. Glad to help you learn something. – xbh Sep 16 '18 at 08:06
  • Would you mind looking at my proof over here; https://math.stackexchange.com/questions/2918350/proving-that-bigcap-n-1-infty-a-n-3-6-unsure-about-correctness-o It is similar to this one and my second try at it is correct I think. Thanks regardless. – Cro-Magnon Sep 16 '18 at 11:56