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.
 A: The whole point of a limit is that you go "up to but not including" the number. The limit of the upper bound of the union you gave is $1$, but that doesn't mean $1$ is in the union. Just like saying $\lim_{x\to 2}f(x)=3$ doesn't imply $f(2)=3$. To be an element of the union, it must be in at least one set. If you can't pick any single set  (pre-union) that contains $1$, then it isn't in the union. It is a limit point, which is where the confusion lies, but that's a different thing entirely.
A: 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 $\varepsilon>0$...? 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 $n>n_0$, $x\in A_n$.
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 "continuous" in the way you would like it to be, that is, $$\lim_{n\to\infty}\chi_{\left[0,1-\frac1n\right]}\neq\chi_{\left[0,1-\lim\limits_{n\to\infty}\frac1n\right]}$$
A: 1 is not in any of the sets, so it can't be in their union.
A: Note that, for this case, (with the "definition" of $\infty$ expanded)
$$\begin{aligned}
\bigcup_{n=1}^\infty \left[0,1-\frac{1}{n}\right] & \stackrel{\operatorname{def}}{=} \lim_{k\to\infty} \bigcup_{n=1}^k \left[0,1-\frac{1}{n}\right] \\
 & = \lim_{k\to\infty} \left[0,1-\frac{1}{k}\right] \\
 & = \left[0,1\right)
\end{aligned}$$
