Difficulty Understanding Rudin's Example 2.10(b) I was studying Rudin's Principles of Mathematical Analysis, Chapter 2 Example 2.10(b) and I came across these two examples which I didn't understand and did not get a clear idea of what they meant:

Let $A$ be the set of a real number $x$ such that $x\in\mathbb{R}_{(0,1]}$. $\forall x\in A$, Let $E_{x}$ be the set of real numbers $y$ such that $y\in\mathbb{R}_{(0,x)}$. Then
$$
\bigcup_{x\in A}E_{x} = E_{1}\qquad \text{and}\qquad \bigcap_{x\in A}E_{x} = \emptyset.
$$

I am having a hard time understanding them and I would be glad if there is a clear proof for these two results.
 A: I think that the fundamental issue here is that the original asker is having difficulty understanding the style and presentation of Rudin's text.  Given that Rudin's style is a fairly common one, it is, I think, worth spending some time to talk about how to read a mathematics text.
In most modern mathematical texts, new ideas are typically introduced via definitions.  If a definition is particularly non-obvious, or if it requires additional clarification, the author will give one or more examples.  Then the author will begin to build a theory by stating one or more theorems (or propositions, or lemmata, or corollaries—essentially, one or more statements which must be proved), and very possibly giving proofs of those results.  This basic outline will be repeated over and over again in the text:  definition, examples, theorem-proof, theorem-proof, repeat.
In Chapter 2 of Principles of Mathematical Analysis, Rudin is building up topological tools which will eventually be used to discuss notions of continuity and differentiability.  In the first section of that chapter, he introduces the reader to some basic ideas from set theory.  For the question asked here, the first relevant part of the text is Definition 2.9 (taken from the 1976 printing—I don't know if the notation changed in later printings or editions):

2.9 Definition  Let $A$ and $\Omega$ be sets, ans suppose that with each element $\alpha$ of $A$ there is associated a subset of $\Omega$ which we denote by $E_{\alpha}$.
...
The union of the sets $E_{\alpha}$ is defined to be the set $S$ such that $x \in S$ if and only if $x\in E_{\alpha}$ for at least one $\alpha\in A$.  We use the notation
$$ S = \bigcup_{\alpha\in A} E_{\alpha}.$$
...
The intersection of the sets $E_{\alpha}$ is defined to be the set $P$ such that $x\in P$ if and only if $x\in E_{\alpha}$ for every $\alpha\in A$.  We use the notation
$$ P = \bigcap_{\alpha\in A} E_{\alpha}.$$

This is a definition.  Rudin is introducing a couple of new concepts (unions and intersections of arbitrary families of sets).  Because new definitions are often opaque, it is generally good practice to give some examples, which Rudin does immediately.  The question here is about the second example:

2.10 Examples (b) Let $A$ be the set of all real numbers $x$ such that $0<x\le 1$.  For every $x \in A$, let $E_x$ be the set of real numbers $y$ such that $0 < y < x$.  Then
$$\text{(ii)}\quad\bigcup_{x\in A} E_x = E_1;
\qquad\text{(iii)}\quad\bigcap_{x\in A} E_x \text{ is empty}. $$

This is an example of a family of sets $\{E_x\}$, indexed by an uncountable set $A$.  Rudin claims that (ii) is "clear"[1], but he does give a short proof of (iii):

...we note that for every $y>0$, $y\not\in E_x$ if $x < y$.  Hence $y \not\in \bigcap_{x\in A} E_x$.

The notation used in this example is kind of funny (as Peter Woolfitt opined, "I'm not quite sure of the pedagogical purpose here for the notation of $A$ and $E_x$..."), but it is meant to match the notation of the previous definition.  Because this example is meant to illustrate the previous definition, one must assume that Rudin finds the example so simple and intuitive that it is worth using to illuminate the definition.[2]  Understanding these examples is meant to give insight into the definition.  So, what does the definition say?
Regarding $S = \bigcup_{x\in A} E_x$, Rudin claims that this is $E_1$.  To show that this is true, we have to show two set inclusions.

*

*First, show that $E_1 \subseteq S$.  Suppose that $y \in E_1$.  Then, by definition of $E_1$,
$$ 0 < y < 1. $$
By definition of the union, we need to show that there is at least one $x \in A$ such that $y \in E_x$.  But
$$ E_x = \{ y : 0 < y < x\}, $$
so any value of $x$ between $y$ and $1$ will get the job done.  For example, take $x = (y+1)/2$.


*Now, show that $E_1 \supseteq S$.  Suppose that $y \in S$.  By definition of the union, there is some $x\in A$ such that $y \in E_x$.  But then
$$0 < y < x. $$
However, $x$ is an element of $A$, and so $x < 1$.  Combining this with the previous inequality,
$$ 0 < y < 1, $$
which means that $y \in E_1$.
Regarding $P = \bigcap_{x\in A} E_x$, Rudin claims that this is empty.  To show that this is the case, it might be helpful to first note that each $E_x$ is a subset of $\Omega = \{z : 0 < z \le 1\}$.  Thus if we want to show that $P$ is empty, we need only show that if $0 < y \le 1$, then there is some $E_x$ such that $y$ is not an element of $E_x$.  So, fix some $y$ with $0 < y \le 1$.  If $x < y$, say $x = y/2$, then
$$ E_x = \{ z : 0 < z < y/2\}. $$
But $y/2 < y$, and so $y \not\in E_x$.  Thus we can find at least one $E_x$ which does not contain $y$, which is sufficient to show that $y\not\in P$.  This holds for every $y$ in our universe, thus $P$ is empty.

[1] A personal pet peeve of mine is when authors claim that anything is clear, or obvious, or trivial.  This kind of language does nothing to further understanding, and only serves to make readers feel dumb.
[2] Personally, I agree with Rudin—I think that this is a good example, as it is (in my opinion) pretty simple once you get your head around the notation.  It is a good example if you want to bang your head against a crunchy definition with opaque notation.
A: Notations: In explicit set-theoretic notations, $E_x$ here means:
$$
E_x := \{y \in \Bbb{R} : 0 < y < x\}
$$
Thus, in particular, we have that:
$$
E_1 := \{y \in \Bbb{R} : 0 < y < 1\}
$$

Proof of statement: I first prove the first statement. Let $y \in \bigcup_{x \in A} E_x$, so $y \in E_x$ for some $x \in A$. In other words, $0 < y < x$ for some $x \in A$. Now since $x \in \Bbb{R}_{(0,1]}$, we have that $x \leq 1$, so $0 < y < 1 \implies y \in E_1$. On the other hand, suppose $y \in E_1 \implies 0 < y < 1$. Observe that $y < \frac{y + 1}{2} < 1$ (can you tell why?), so $y \in E_\frac{y+1}{2} \subseteq \bigcup_{x \in A} E_x$.
For the second statement, we prove by contradiction. Suppose $\bigcap_{x \in A} E_x$ is non-empty, so let $y \in \bigcap_{x \in A} E_x$. Since $y > 0$, we have that $0 < \frac{y}{2} < y$, so we can't have $y \in E_\frac{y}{2}$ (i.e. $y \notin E_\frac{y}{2}$). Yet, by definition of intersection, we have that $\bigcap_{x \in A} E_x \subseteq E_\frac{y}{2}$, a contradiction.
A: I'm not quite sure of the pedagogical purpose here for the notation of $A$ and $E_x$, but $A$ can be written as a half-open interval $A=(0,1]$ and $E_x$ can be written as the open interval $(0,x)$.
Therefore
$$\bigcup_{x\in A}E_{x} =\bigcup_{x\in (0,1]}(0,x)= (0,1)=E_1$$
Note this is true because the set $(0,1)$ is in the union and $(0,x)\subseteq(0,1)$ for all $x\in (0,1]$.
On the other hand, we can show
$$\bigcap_{x\in A}E_{x} =\bigcap_{x\in (0,1]}(0,x)= \emptyset$$
by contradiction.
If some real number $a$ is in this intersection, then $a\in(0,1)$ because every set involved in the intersection is a subset of $(0,1)$. However, this means that the intersection uses the set $(0,a)$ which does not actually include $a$ itself. Therefore $a$ cannot be in the intersection, so the intersection is empty.
A: Hint:
In normal set notation, $A=(0,1]$ and $E_x=(0,x)$.
Prove $E_x\subseteq E_y \iff x\le y$.
A: My copy of Rudin words it differently.  But the point is, it's an example of the notation that if you have a master set $A$ and for each element $x \in A$ there is a subset of $A$ that is somehow defined and associated with the element $x$ we use the notation $E_x \subset A$.
In this example $A = (0,1]$  and for any $x \in A$ we define $E_x = (0, x)$.
We are using the notation that because $x \in A$ and $x$ defines our subset $(0,x)$ we use that notation $E_x$ and that subset that is defined be $x$ as in $E_x =\{y \in A|y < x\}=(0,x)$.  For a different element $w\in A$ we'd have a different subset $E_w$ defined by $w$.  So $E_{\frac 12} = (0,\frac 12)$ and $E_{0.879356359} = (0, 0.879356359)$ then notation "E sub something" means that the "something" was "key" in defining what the subset was.
The statement $\cup_{x\in A} E_x = E_1$ simply means $\cup_{x\in A = (0,1]} (0,x) = (0,1)$.  I'm going to assume you don't need that proven.
And the statement $\cap_{x\in A} E_x =\emptyset$ simply means $\cap_{x\in A=(0,1]} (0,x) = \emptyset$.  Ditto.
====
Actually working through a proof would be a good idea to get used to the concepts attempting to be taught.
$\cup_{x\in A} E_x = \{y\in A|$ so that $y\in E_x$ for some $x \in A\}=$
$\{y \in A|$ so that there is an $x\in A$ so that $0< y < x\}=$
$\{y\in \mathbb R|$ so that $0< y \le 1$ and there is an $x$ so that $0<x \le 1$ so that $0< y < x\}=$
$\{y\in \mathbb R|$ so that there is an $x$ so that $0 < y < x < 1\}=$
$\{y \in \mathbb R$ so that $y \in (0,1]$ but the is another number $x \in (0,1]$ so that $y < x\}=$

but that is all numbers in $(0,1)$ except $1$.  If $0 < y < 1$ then there is an $x $ so that $y < x  <1$ so $y$ can be any $y\in(0,1)$ but $y \ne 1$ as there is no $x \in (0,1]$ so that $1 < x \le 1$ (that would imply $1< 1$).

$\{y \in \mathbb R|$ so that $0 < y < 1\}$.
Now $E_1$ is, by definition, $E_1 = \{y\in (0,1]| 0< y < 1\}=(0,1) \subset (0,1]\subset \mathbb R$, we conclude $E_1 = \{y\in \mathbb R| 0< y< 1\}$ which, by above is $\cup_{x\in A}E_x$.
And $\cap_{x\in A}E_x =\{y \in A|y \in Ex$ for all $x \in A\}=$
$\{y\in \mathbb R| 0< y \le 1$ and for every $x$ so that $0< x \le 1$ we have $y$ is one of the real numbers $w$ so that $0 < w < x\}=$
$\{y\in \mathbb R| $ so that for every $x\in (0,1]$ we have $0< y < x\}=$

But  there certainly are no such $y$ as if there were, we would have $y\in (0,1]$ and so $0 < y < y$ which is not possible.

So $=\emptyset$.
