Accumulation points with respect to Zariski Topology on $\mathbb{R}$.

Question

Consider the topology on $\mathbb{R}$ : the closed sets are given by the finite sets and $\mathbb{R}$ (basically the Zariski Topology on $\mathbb{R}$).

Let $(x_n)_{n \in \mathbb{N} }$ be a sequence in $\mathbb{R}$.

1. How do we show : if $l$ is a limit of the sequence $(x_n)_{n \in \mathbb{N} }$ if and only if for all $x \neq l$, the number of indices $n$ such that $x_n = x$ is finite.

2. Determine the accumulation points of the sequence $(x_n)_{n \in \mathbb{N} }$ w.r.t. Zariski topology

Answer 1

Suppose that $\lim x_n=l$ and $x\neq l$. $U_x=\mathbb{R}-\{x\}$ is an open subset which contains $l$, there exists $N$ such that $n>N$ implies that $x_n\in U_x$. This implies that $x_n\neq x$ for $n>N$.

Conversely suppose that $x\neq l$ implies that the set of $x_n$ such that $x_n\neq x$ is finite. Let $U$ be an open subset containing $l$, $U=\mathbb{R}-\{u_1,...,u_p\}$. Let $A\subset \mathbb{N}$ such that $a\in A$ implies that $x_a=x_i$. $A$ is finite. Let $N=\max A$. For every $n>N$, $x_n\neq u_i$ implies that $x_n\in U$.

Answer 2

You can translate the definition of “being a limit” using closed sets:

$ $          for every closed set $C$ not containing $l$, there exists $m$ such that, for $n>m$, $x_n\notin C$

The closed sets non containing $l$ are necessarily finite.

Suppose $l$ is a limit of the sequence. Then, taking $C=\{x\}$, with $x\ne l$, we see that the set $\{n\in\mathbb{N}:x_n=x\}$ is upper bounded, hence finite.

Suppose $l$ satisfies the condition about finiteness. Take a closed set $C$ not containing $l$. Then $C=\{r_1,r_2,\dots,r_k\}$ is finite and you should be able to finish up.

What about accumulation points of the sequence? Using Kelley's terminology, a sequence is convergent to $l$ if (and only if) it is eventually in every (open) neighborhood of $l$ (it belongs there from a certain point on). The point $l$ is an accumulation point if (and only if) it is frequently in every (open) neighborhood of $l$ (for every $m$, there is $n$ such that $n>m$ and $x_n$ is in the given neighborhood).

If $l$ is an accumulation point of the sequence and $x\ne l$, for every $m$ there is $n>m$ such that $x_n\ne x$. In particular, $\{n\in\mathbb{N}:x_n\ne x\}$ is infinite.

Can you finish with the converse?