Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My textbook General Topology by Pervin) says "If $x$ is a point and $E$ a subset of a $T_{1}$-space $X$ satisfying the first axiom of countability, then $x$ is a limit point of $E$ iff there exists a sequence of distinct points in $E$ converging to $X$"?

I don't understand why $X$ needs to be a $T_{1}$-space for this to be true.

Let $X$ be a first axiom topological space, and $E\subseteq X$. If a point $a\in X$ is a limit point of $E$, then there is a sequence in $E$ whose limit is $a$. Proof: let $B_{n}(a)$ be a monotone decreasing countable open base containing $a$. All these sets will contain at least one point from $E$. If we keep on choosing a countable number of points from each $B_{i}(a)\cap E$, we will generate one such sequence converging to $a$.

It is also clearly true that if there is a sequence in $E$, which converges to $a$, then $a$ is a limit point of $E$.

Nowehere in this argument was the fact that $E$ is a $T_{1}$-space is used.

I feel $X$ being a $T_{1}$-space only proves that $\bigcap_{i}B_{i}(a)=\{a\}$. This fact I feel is irrelevant in making the argument concerned.

Thank you for your time.

share|improve this question
    
Your text claims something more than you prove, namely that there is a sequence of distinct points converging to $x$. –  Miha Habič Jun 16 '13 at 10:18
    
I don't see why that should not be possible in my argument. –  Ayush Khaitan Jun 16 '13 at 10:20
1  
Suppose that any neighbourhood of $x$ contains some point $y\in E$ and that $\{x,y\}$ is open. You cannot possibly get an injective sequence to converge to $x$ in this case. –  Miha Habič Jun 16 '13 at 10:31
    
We know $y\in E$. $\{x,y\}$ has to be equal to some $B_{i}(x)$, and all $B_{j}(x)=B_{i}(x)$ for $j>i, i,j\in\mathbb{N}$ (as $x$ is a limit point of $E$). By selecting $y$ from every $B_{j}$ for $j>i$, I'm stil creating a sequence converging to $x$!! This would not be true if the statement said "infinite distinct points". However, this is not what the statement says. –  Ayush Khaitan Jun 16 '13 at 10:54
1  
As I (and I think most people) understand the statement, it is asserting that in a first countable $T_1$ space, a point is a limit point of $E$ if there is an injective sequence in $E$ converging to it. I do agree that first countability suffices to get some (not necessarily injective) sequence. –  Miha Habič Jun 16 '13 at 11:07

1 Answer 1

up vote 4 down vote accepted

The key word here is "distinct". So let $X$ be a fixed $T_1$-space. Let $E \subset X$.

First note that a limit point of $E$ by the standard definition is a point $x$ such that every (open) neighbourhood $U$ of $x$ contains a point of $E$ different from $x$.

Now, for a $T_1$-space we can say something stronger: $x$ is a limit point of $E$ iff every (open) neighbourhood of $x$ contains infinitely many points from $E$.

Proof: if every neighbourhood contains infinitely many points of $E$, it will certainly contain one different from $x$, so one implication is trivial, so to see the other one: suppose a limit point $x$ of $E$ has an open neighbourhood $U$ such that $ U \cap E$ is finite. Then $F = (U \cap E) \setminus \{x\}$ is also finite, and thus a closed set (here we use $T_1$: singletons and thus finite sets are closed), and so $X \setminus F$ is open, and so is $U' = U \cap (X \setminus F)$. As the latter set is thus an open neighbourhood of $x$ such that $U' \cap E \subset \{x\}$ (i.e. either empty or just $x$, depending on whether $x \in E$ or not), this contradicts that $x$ is a limit point of $E$.

This is the reason the fact is stated for $T_1$-spaces. In a non-$T_1$ space like Sierpinski space $\{0,1\}$ with open sets $\{\emptyset, \{0,1\},\{0\}\}$, which is $T_0$, the point $1$ is a limit point of the finite set $E = \{0\}$, but we certainly cannot find infinitely many different points from $E$ converging to $1$, even though the space is trivially first countable (and the constant sequence of $0$'s does converge to $1$).

Now if there is a sequence of distinct points from $E$ that converges to $x$, then clearly $x$ is a limit point of $E$ (every open neighbourhood of $x$ contains a tail of the sequence, and so even infinitely many different points of $E$, so we get the strong version of limit point for free). This holds without $T_1$ or first countability.

Now if $X$ is $T_1$ and also first countable, and if $x$ is a limit point, then let $U_n$ be a local base at $x$, and pick $x_1 \in E \cap U_1$. When we have chosen $x_1,\ldots,x_n$ (all distinct and from $E$) such that $x_i \in U_j$ for all $j \le i$ ( for $i=1 \ldots n$), then pick $x_{n+1} \in E \cap (U_1 \cap \ldots \cap U_n \cap U_{n+1})$, such that $x_{n+1}$ is distinct from all $x_1,\ldots,x_n$. This can be done as $U_1 \cap \ldots U_n \cap U_{n+1}$ is an open neighbourhood of $x$ and every one of them intersects $E$ in infinitely many different points by the strengthened version of limit point, which is valid in $T_1$-spaces. This defines a sequence by recursion, and it clearly converges to $x$ and consists of all distinct points from $E$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.