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I don't know how can I make a correct proof about the following statement:

$T_{n}= [0,1-\frac{1}{n}]$

$\bigcup_{n\in \mathbb{N}}^{ } T_{n}=[0,1)$

I understand the logic, but I don't know how can I prove it in a formal way.

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    You prove inclusion both ways. Take a real number $x \in [0,1)$. How do you prove it's in the union of the $T_n$? The other direction is also easy, maybe easier, but tackle one direction at a time. – 2017-01-19

2 Answers 2

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Since $T_n\subset[0,1)$ for every $n\in\Bbb N$, then $\bigcup_{n\in\Bbb N}T_n\subset[0,1)$. This is trivial. Let us prove the converse inclusion. To this end take $x\in[0,1)$. Bacause $\lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)=1$, there exists $n\in\Bbb N$ s.t. $0\le x\le 1-\dfrac{1}{n}$. Hence $x\in T_n$, so $x\in \bigcup_{n\in\Bbb N}T_n$.

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    But, we are trying to prove x < 1 - (1 / n) and not x ≤ 1 - (1 / n)? Why in this way? – 2017-01-19
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We can show that $\cup_{n \in \mathbb{N}}T_n$ is a subset of $[0,1)$ and vice versa, as Fabio Somenzi suggests in his comment.

First, let $x \in \cup_{n \in \mathbb{N}}T_n$. Thus there exists an $n$ such that $x \in T_n$, i.e. such that $x \leq 1 - \frac{1}{n}$. This implies that $x < 1$, so that $x \in [0,1)$.

Conversely, let $x \in [0,1)$. Then $x < 1$ and there exists an $n$ large enough such that $x < 1 - \frac{1}{n}$. For this $n$, we have $x \in T_n$, so that $x \in \cup_{n \in \mathbb{N}} T_n$.

Therefore the sets are contained in each other and must be equal.

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    Why $x \leq 1 - \frac{1}{n}$ implies $ x < 1 $ ? Is it becuase, we are always going to obtain a number lesser than 1? – 2017-01-19
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    yes. $x \leq 1 - \frac{1}{n} < 1$. – 2017-01-19
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    And why if $ \lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)=1 $? – 2017-01-19
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    because as $n \to \infty$, $\frac{1}{n} \to 0$ so $1-\frac{1}{n} \to 1$. Thus by the definition of a limit you can find an $n$ so large that $x < 1 - \frac{1}{n}$, since $x$ is strictly less than 1, and $1-\frac{1}{n}$ can become arbitrarily close to 1. – 2017-01-19
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    And why in the second part of the proof, it's not strictly less tan (1 - 1 / n)? – 2017-01-19
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    it doesn't matter. if you find an $n$ such that $x = 1-1/n$ then just take $n+1$ and you will have found a number for which strict inequality holds. – 2017-01-19