0

As I understand it, he is defining the set $\{q_1, q_2, ... , q_n\}$ as the image of a bijective function whose domain $D$ is the intersection of $N$ and $E$. My question is: How we can verify that this image can exist as a subset of the same metric space $X$ (with $p$, $N$, and $E$) so that $d(p,q_i)$ is even defined? theorem 2.20 in rudintheorem 2.20 pt 2

edit: enter image description here

thankfulperson
  • 113
  • 1
    What bijective function? Do you mean his $d$? That's the metric. – Randall Jul 25 '19 at 17:01
  • 2
    You're introducing a bijective function that's not in the quoted text from Rudin. The text says $N\cap E-{p}={q_1,\dots,q_n}$. – Andreas Blass Jul 25 '19 at 17:12
  • 1
    This may help: https://math.stackexchange.com/q/509241/597411 – C. Melton Jul 25 '19 at 17:14
  • I though, looking at definition 2.7 in the text, writing any at most countable set as a sequence of elements is defined as the result of a bijective function. I can attach the image of 2.7 in the main post if you don't have the text. – thankfulperson Jul 25 '19 at 17:25
  • for instance if the set was something like {{q1, q2}, q3}, this is not “equal” to {q1, q2, q3} but only equivalent. – thankfulperson Jul 25 '19 at 17:33
  • 1
    The domain of the bijective function in the definition of sequence is the set of positive integers. We just write $q_1$ instead of $q(1)$, $q_2$ instead of $q(2)$ and so on. All the points discussed in the theorem are elements of the same metric space, so there is no question that $d(p,q_m)$ is defined. – saulspatz Jul 25 '19 at 17:37

1 Answers1

1

The definition of limit point is as follows:

$$\text{$p\in E$ such that for all $\epsilon>0$, there is $q\in E\backslash\{p\}$ such that $d(p,q)<\epsilon$}$$

Let $X$ be a metric space. Suppose $p$ is a limit point of $E\subseteq X$. Suppose that for some $r>0$ there is $N=\{x\in E: d(p,x)<r\}$ such that $(N\cap E)\backslash\{p\}=\{q_1,q_2,...,q_n\}=A$.

$A$ is simply the intersection between two sets with one element (namely $p$) removed; there are no functions involved. If you wish, you can define $f:A\to A$ such that $f(x)=x$, but this really isn't necessary.

$X$ is a metric space, so the metric $d$ on $X$ exists by definition. Since $A \subseteq X$ and $p\in E \subseteq X$, $d(p,q_i)$ is defined for all $q_i\in A$.

C. Melton
  • 338
  • I get it now. Basically each qi is an element of the intersection of N and E (a subset of X) and therefore obviously an element of X, regardless of how the elements are/were grouped/listed. I honestly don’t know why I didn’t recognize this. Thanks! – thankfulperson Jul 25 '19 at 18:07
  • @CadeBruce Exactly. I'm glad it makes sense to you now. This is how we learn! – C. Melton Jul 25 '19 at 18:14