I'm new at the mathematic analysis and currently reading Zorich's book. When I read the definition about the limit point, I noticed that it's a bit different than Rudin's "Mathematical Principle" Rudin's definition is

A point $p$ $\in$ $R$ is a limit point of $X\subset R$ if every neiborhood of the point $p$ contains an at least one point different from $p$ itself.

While the Zorich's definition is

A point $p$ $\in$ $R$ is a limit point of $X\subset R$ if every neiborhood of the point $p$ contains an infinite subset of $X$.

The book of Zorich then mentions Rudin's definition and says these two are equivalent and asks for verification. Here is what I tried.

From Zorich's definition, $\forall N(p),\exists S\subset X(S\ is\ infinite)$, the Rudin's definition is quite obvious, just find another point in $S$

Here is what really have confused me.

From Rudin's definition, I begin with finding a subset $S$ of $N(p)$ that contains $p$, and then try to prove such $S$ can not be finite by contradiction.

Here is what I did, Let $S$ be a subset of $X$ such that $q\in S\subset X$, then assume $S$ is finite. Since $S$ is finite, there exists such $q^\prime\in S$ such that $d(q^\prime,p)=d(q,p)_{min}$, Then for the neiborhood of $p$ with radius $r<d(q^\prime,p)$, there is no such $q\in N(p)$ exists then contradict.

But then I realize that my proof is some sort of prove for a $S$ which is the set containing all $q$, not for some specific $q$. Since there might be infinite $q$, my proof is wrong.(from my understading)

Is it right or wrong, and why. I am really new at this. And it's my first question in this site, if I did something wrong pls tell me. THX.

  • $\begingroup$ Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is. $\endgroup$ – xbh Jan 23 '19 at 12:34
  • $\begingroup$ Tips on typing math: to use the prime $'$, just use the single quotation mark ', or use the command \prime by typing something like p^\prime, as in $p^\prime$. $\endgroup$ – xbh Jan 23 '19 at 12:37
  • $\begingroup$ THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$. $\endgroup$ – 卢弘毅 Jan 23 '19 at 12:40
  • $\begingroup$ These are equivalent over any Hausdorff Space. Rudin's definition is more general though. $\endgroup$ – Brevan Ellefsen Jan 23 '19 at 21:40

I don't understand your proof, and therefore I cannot comment on it. However, if $p\in\mathbb R$ is such that some neighborhood $U$ of $p$ only has finitely many elements of $X$, then $p$ is not a limit point of $X$. In fact, let $F=U\cap X$. I am assuming that it is finite. Take $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\cap F=\emptyset$. Then $U\cap(x-\varepsilon,x+\varepsilon)$ is a neighborhood of $x$ which has no element of $X$.

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  • $\begingroup$ Thanks a lot, I just edited my $S$ as a set that contains all possible $q$ $\endgroup$ – 卢弘毅 Jan 23 '19 at 12:42
  • $\begingroup$ I still don't get your definition of $S$. You wrote “Let $S$ be a subset of $X$ such that $q\in S\subset X$”. Since you wrote before that $S$ is a subset of $X$, the final two characters are redundant. But what is $q$? $\endgroup$ – José Carlos Santos Jan 23 '19 at 13:10

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