Is (Baby) Rudin being lazy in Theorem 2.38? (I.e., assuming something holds for all sets in an infinite intersection is sufficient) The theorem in question is: If $\{I_n\}$ is a sequence of intervals on $\mathbb{R}^1$ such that $I_n\supset I_{n+1}$, then $\cap_{n=1}^\infty I_n\neq\emptyset.$
Rudin says to let $I_n=[a_n,b_n]$, and since $\{a_n\}$ is bounded above by $b_1$, it has a sup, which we can call $x$. $a_m \leq x\leq b_m$, so $x\in I_m$ for all $m=1,2,3,...$, so $x\in\cap_{n=1}^\infty I_n\Rightarrow\cap_{n=1}^\infty I_n\neq\emptyset.$
This strikes me as a bit odd: Rudin says something about every set in an infinite intersection and doesn't really worry about the potential qualitative difference between each object in the intersection and the intersection itself--e.g., each interval has a nonzero length, but the intersection may not.
I suppose it doesn't matter here because membership doesn't seem to change qualitatively in the same way length does. Nevertheless, it does seem a bit 'sloppy' in a way to make the final deduction Rudin does. Or should it be clear on a case by case basis when this 'jump' is valid to make?
It seems to me less 'sloppy' to show that $\cap_{n=1}^\infty I_n \supset [\alpha,\beta],$ where $\alpha=\sup\{a_n\}$ and $\beta=\inf\{b_n\}$. And then show $\alpha\leq\beta$. Since $\alpha\leq\beta$, $[\alpha,\beta]\neq\emptyset$, $\cap_{n=1}^\infty I_n\neq\emptyset.$ But perhaps the same 'jump' I accuse Rudin of making is still here, just better hidden from me...
 A: It is not sloppy to say that if $x \in I_m$ for all $m$, then $$x \in \bigcap_{m=1}^\infty I_m.$$ This is simply the definition of the infinite intersection: an element is in $\bigcap_{m=1}^\infty I_m$ if and only if it is in $I_m$ for every $m$.
It is in some sense more "sloppy" to say that just because $[\alpha, \beta] \subseteq I_m$ for all $m$, then $[\alpha,\beta] \subseteq \bigcap_{m=1}^\infty I_m$. It is still true, but it does not follow directly from the definition. To prove it, we'd...
...well, we'd take an arbitrary $x \in [\alpha, \beta]$, use $[\alpha, \beta] \subseteq I_m$ to conclude that $x \in I_m$ for all $m$, and then use the definition of $\bigcap$ to conclude that $x \in \bigcap_{m=1}^\infty I_m$.
A: I see no sloppiness: Rudin leaves something to be proved as an exercise.
Consider the set $A=\{a_n:n\ge1\}$, the set of left boundaries of the given intervals.
Since $a_n\le b_1$ for every $n$, we have that $A$ is bounded so it has a supremum $x$. By definition, $x\ge a_n$, for every $n$. But we can also prove that $x\le b_n$, for every $n$. Indeed, not only $b_1$ is an upper bound of $A$, but every $b_m$ is an upper bound as well: suppose not; then we would have $a_n>b_m$, for some $n$, but taking $k$ the maximum between $n$ and $m$, we have $a_n\le a_k\le b_k\le b_m$, a contradiction.
Therefore $x\le b_n$, for every $n$, because it is the least upper bound of $A$. Together with $x\ge a_n$, for every $n$, we obtain $x\in[a_n,b_n]$, for every $n$, which is the same as saying that
$$
x\in\bigcap_{n\ge1}[a_n,b_n]
$$
You can prove that the intersection has the form $[x,y]$, where $y=\inf\{b_n:n\ge1\}$, but it's not needed. Note that it can be $x=y$, which is actually the most useful case when applying the theorem.
