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I've been looking around (websites, notes and so on) but I never found a proof for this:

$$\bigcup_{n = 1}^{+\infty} \left[\frac{1}{n}, 1 - \frac{1}{n}\right] = (0, 1)$$

I am not understanding why though. If I think about, this is just

$$[1, 0] \cup \{\frac{1}{2} \} \cup [\frac{1}{3}, \frac{2}{3}] \cup \ldots $$

And for $n\to +\infty$ we reach $(0, 1)$

But $0$ belongs to the set, so I would say the union is $[0, 1)$.

I can't get why it's $(0, 1)$.

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    $\begingroup$ $[1,0]$ is empty because no $x$ satisfy $1 < x < 0$ $\endgroup$
    – CyclotomicField
    Commented Nov 20, 2023 at 21:42

1 Answer 1

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For $n = 1$ the interval in the union is $[1,0]$. Do not confuse this with $[0,1]$, which is something different! By definition, $$[1,0] = \{x \in \mathbb{R} : 1 \leq x \leq 0\} = \emptyset,$$ since no $x$ can fulfill the requirement to be in this set. The case $n = 1$ can therefore be left out from the union. It would probably have been better style to start at $n = 2$ as intervals with inverted limits are confusing.

You will now notice that both $0$ and $1$ are contained in none of the intervals in the union. Therefore, they are not contained in the union itself.

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  • $\begingroup$ Ah! Gosh, I completely forgot about the order relation!! $\endgroup$
    – Heidegger
    Commented Nov 20, 2023 at 21:45
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    $\begingroup$ +1 would have been better $\endgroup$
    – GEdgar
    Commented Nov 20, 2023 at 21:47

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