Proving $\bigcup_{n=1}^{\infty}[3 - \frac{1}{n}, 6] = [2, 6]$, have I done it correctly?

Question

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.

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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]$

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

Answer 1

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 $$[2,6]=\left[3-\frac{1}{1},6\right]\subseteq\bigcup_{n=1}^\infty\left[3-\frac{1}{n},6\right].$$

Answer 2

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 $1$.

Answer 3

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 $$ \bigcup_1^\infty \left[3 - \frac 1n, 6\right] = [2,6]. \blacktriangleright $$