# Some basic questions about two definition of limit point in Zorich‘s “Mathematical Analysis I”

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 $$r, there is no such $$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.

• Sorry, but I am not quite sure about the usage of the set $S$. I cannot get what the set $S$ exactly is. – xbh Jan 23 '19 at 12:34
• 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 $p^\prime$. – xbh Jan 23 '19 at 12:37
• THXs for the tip, I just rewrote the usage of $S$ as the set which contains all $q$. – 卢弘毅 Jan 23 '19 at 12:40
• These are equivalent over any Hausdorff Space. Rudin's definition is more general though. – 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 $$X$$.
• Thanks a lot, I just edited my $S$ as a set that contains all possible $q$ – 卢弘毅 Jan 23 '19 at 12:42
• 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 $q$? – José Carlos Santos Jan 23 '19 at 13:10