What is $\bigcup\limits_{n=1}^\infty [0,1-\frac{1}{n}]$? This is probably a pretty dumb question, but I am confused by set theory again. The question is whether 
$$\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right]$$
equals $[0,1]$ or $[0,1)$. However, I am looking for some explanation and not only the result, since I'd like to understand why it's the one or the other.
Remember that $x\in\bigcup\limits_{i\in I} A_i$ if and only if for some $i\in I$, $x\in A_i$.
So $1$ is in the union if and only if it appears in at least on of those intervals, so is it? No. It is not.

To the comment, don't think of $\bigcup_{i=1}^\infty$ as a limit in the calculus-sense of the word. Think of it as a logical operation which tells you that the index set is $\mathbb N$ (or some other set which is clear from context) and then use the above formula.
If you wish to think about it as $f(n)=\bigcup\limits_{k=1}^n [0,1-\frac1k]$, and think about the infinite union as $\lim\limits_{n\to\infty} f(n)$, then there are several caveats:

Limits will usually require some sort of topology, some underlying structure which tells us about convergence. How would you define the limit here? For every $\varepsilon&gt;0$...? It makes no sense, since subsets of $\mathbb R$ do not have a natural metric function.

We can consider the following definition: $A$ is the limit of the sequence of $f(n)$ if and only if for every $x\in A$, there exists $n_0$ such that for all $n&gt;n_0$, $x\in A_n$.
Observe, however, that this coincides with the definition above, that $x$ is in the union if and only if it appears in at least one of the functions. This definition, however, coincides with the above only because this sequence of sets is increasing.

Luckily, we can always think of an infinite union as an increasing sequence, but we would expect a definition for a limit to work for any sequence of sets, not just increasing unions.

We can, however, think of it as a limit of a sequence of characteristic functions, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}=\chi_{[0,1)}$$
even as such limit, though, it is not &quot;continuous&quot; in the way you would like it to be, that is, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}\neq\chi_{\left[0,1-\lim\limits_{n\to\infty}\frac1n\right]}$$
Note that, for this case, (with the "definition" of $\infty$ expanded)

$$\begin{aligned}
\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right] &amp; \stackrel{\operatorname{def}}{=} \lim_{k\to\infty} \bigcup_{n=1}^k \left[0,1-\frac{1}{n}\right] \\
 &amp; = \lim_{k\to\infty} \left[0,1-\frac{1}{k}\right] \\
 &amp; = \left[0,1\right)
\end{aligned}$$