# 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 . Then

(i) $$E_{x} \subset E_{z}$$ if and only if $$0 ;

(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 $$x. Hence $$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.

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 $$0. This is the same as the open interval $$(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

(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 $$0, just because that's the way we defined our set $$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!