关于实数集有限补拓扑下极限点的疑问（基于Armstrong拓扑学示例）

Hey there! Let's unpack this confusion about limit points in the finite complement (cofinite) topology—this is a classic tricky spot when first learning topology, so you're not alone here.

First, let's recap the key definitions we need to make this click:

1. Limit Point Definition #

A point $x \in X$ is a limit point of a subset $A \subseteq X$ if every open set containing $x$ intersects $A$ at some point other than $x$. In formal terms:

For every open set $U$ with $x \in U$, $U \cap (A \setminus {x}) \neq \emptyset$.

2. Finite Complement Topology Rules #

On $X = \mathbb{R}$, the finite complement topology defines open sets as:

• The entire set $\mathbb{R}$ itself, or
• Any subset $U \subseteq \mathbb{R}$ where the complement $X \setminus U$ is a finite set (i.e., $U$ is $\mathbb{R}$ minus a finite number of points).

Why Every Point in $\mathbb{R}$ is a Limit Point of an Infinite Subset $A$ #

Let's take your example: $A = \mathbb{Z}$ (the integers, an infinite subset of $\mathbb{R}$). Pick any arbitrary point $x \in \mathbb{R}$—we need to show $x$ is a limit point of $A$.

Here's the breakdown:

• Take any open set $U$ that contains $x$. By the finite complement topology rules, $X \setminus U$ is either empty (if $U = \mathbb{R}$) or a finite set of points.
• Now look at $A \setminus {x}$: since $A$ is infinite, removing just one point ($x$) leaves it still infinite.
• When we take $U \cap (A \setminus {x})$, this is equivalent to taking the infinite set $A \setminus {x}$ and removing the finite number of points that are in $X \setminus U$. An infinite set minus a finite number of points is still infinite (and definitely non-empty!).
• That means $U \cap (A \setminus {x})$ can never be empty—no matter which open set $U$ we pick containing $x$, it will always contain other points from $A$.

Concrete Example #

Let's make this tangible:

• Let $x = 0.5$ (not an integer), and take an open set $U = \mathbb{R} \setminus {1, 2, 3}$. This $U$ contains $0.5$, and $U \cap (\mathbb{Z} \setminus {0.5}) = \mathbb{Z} \setminus {1,2,3}$—which includes 0, -1, 4, etc. Clearly non-empty.
• Now take $x = 1$ (which is in $A$), and an open set $U = \mathbb{R} \setminus {2, 4}$. $U$ contains 1, and $U \cap (\mathbb{Z} \setminus {1}) = \mathbb{Z} \setminus {1,2,4}$—still full of points like 0, -1, 3, etc.

The core idea here is that finite complement open sets can only exclude a finite number of points, but your subset $A$ is infinite. No matter how many finite points you exclude, there will always be plenty of $A$ left in the open set to satisfy the limit condition.

内容的提问来源于stack exchange，提问作者innumeratus