Problem in Understanding Baby Rudin : Chapter 2 : Basic Topology, page number : 28, 2.10(b) I'm reading the "Baby Rudin" book and one particular example (which is very trivial, I guess but) I couldn't understand properly. I'm stating the problem and the answer here.
I want the visual and logical understanding of this problem.
[Chapter 2 : Basic Topology, page number : 28, 2.10(b)] :

Let $A$ be the set of real numbers such that $0 < x \leq 1$. Fro every
$x \in A$, let $E_{x}$ be the set of real numbers $y$ such that $0 <y <x$. Then
(i) $E_{x} \subset E_{z}$ if and only if $ 0 <x \leq z \leq 1$;
(ii) $\bigcup_{x \in A} E_{x} = E_{1}$;
(iii) $\bigcap_{x \in A} E_{x}$ is empty;
(i) and (ii) are clear. To prove (iii), we note that for every $y>0$,
$y \notin E_{x}$ if $x<y$. Hence $y \notin \bigcap_{x \in A} E_{x}$.

My (wrong?) undertsanding :
(i) From the concept of equality it's obvious. Am I right? What's the proper explanation of this?
(ii) How it's $\bigcup_{x \in A} E_{x} = E_{1}$? Didn't understand.
(iii) For this one with the given hint, I tried to think as points in the real line $x$, but I can't convince myself enough with imagination and logic.
P.S : There are already some answers on this same topic (but not on all 3 points) , but I need the answers in more details in all 3 points, like visually with some sketches in real line, I hope this question will not get duplicated.
 A: Let me try to offer some explanation, unfortunately without sketches, but I hope it will help nonetheless.
(i) The set $E_x$ are all real numbers $y$ such that $0<y<x$. This is the same as the open interval $(0,x),$ so it's really easy to draw: If, say, $x = \frac{1}{2}$ then it's just the line from $0$ to $\frac{1}{2}$ (without the endpoints).
Now the statement of (i) may be more clear: When is a line from 0 to $x$ part of the line from 0 to $z$? Well, the line to $z$ must be at least as long, i.e. $x\leq z$ must be true. But of course $x$ and $z$ must be between 0 and 1 (by the definition of the sets $E_x$ and $E_z$), so finally we get:
$$ E_x \subseteq E_z \iff 0<x\leq z\leq1$$
(ii) Let now $x = 1$, so $E_x = E_1$ is the open interval $(0,1)$. For any other $z \in A$, the open interval $(0,z) = E_z \subseteq E_1$ by (i). Now if we take the union $\bigcup_{x \in A}E_x$ there is not much happening: since $E_1$ is in this union, and all other $E_z \subseteq E_1$, the union is just going to end up to be $E_1$ itself. You simply don't add any points to $E_1$ that are not already in $E_1$.
(iii) To see that the intersection is empty, it is maybe easiest to 'try' to put an element into the intersection and see that it is impossible. What I mean by that is the following:
Let's say $\bigcap_{x \in A} E_x$ is not empty, so it contains some number $y$. We have that $0<y$, just because that's the way we defined our set $A$ (it doesn't include the point 0). Then let us define $x = \frac{y}{2}$. But now, the point $y$ is not in the interval $(0,x) = E_x$ because it is simply too big. (You can think of an explicit example, like $\frac{1}{2}$ is not in the interval $(0, \frac{1}{4})$.)
But $y \in \bigcap_{x \in A} E_x$ means that $y$ must be in all of the $E_x$, which it is not since we explicitly found one that does not contain $y$. So contrary to our assumption, $y \notin \bigcap_{x \in A} E_x$. Since we chose $y$ arbitrarily, this shows that no point can be in $\bigcap_{x \in A} E_x$ which therefore must be empty.
I hope this sheds some light into the matter!
