A condensation point is defined in Exercise (2.)27 on Page 45 of Walter Rudin's Principles of Mathematical Analysis, 3rd Edition as follows:
Define a point $p$ in a metric space $\mathbb X$ to be a condensation point of a set $\mathbb E \subset \mathbb X$ if every neighborhood of $p$ contains uncountably many points of $\mathbb E$.
In light of this definition, the following question came to my mind:
Let $\mathbb X$ be a metric space consisting of all ordered pairs $(\alpha_1, \alpha_2)$ where $\alpha_1, \alpha_2 \in \mathbb Q$ (rational numbers). Let $\mathbb E$ be any subset of $\mathbb X$. Does $\mathbb E$ have a condensation point?
I tend to think that the answer is "no". That's because, $\mathbb X = \mathbb Q \times \mathbb Q$ is countable $\implies$ each neighborhood of each point $p$ in $\mathbb E$ does not include uncountably many points of $\mathbb E$ $\implies$ $p$ fails to satisfy the requirement of being a condensation point of $\mathbb E$. In other words, no subset of $\mathbb Q \times \mathbb Q$ has a condensation point. I'm not sure somehow if this argument is correct, and would appreciate help. Thanks.
Best Answer
You're right: in a countable space $\Bbb X$ there are no condensation points, for any subset, for the simple cardinality reason that all subsets of $\Bbb X$ are countable.
There is a more "cardinal-neutral" definition: if $A$ is an infinite subset of $\Bbb X$ and $x \in X$, then $x$ is called a point of total accumulation of $A$ iff every neighbourhood $U$ of $x$ satisfies $|A \cap U| = |A|$ (so is as infinite as it can be, given the size of $A$). Then in $\Bbb X = \Bbb Q \times \Bbb Q$ all points are points of total accumulation of $\Bbb X$. Fun fact: $\Bbb X$ is compact iff every infinite set $A \subseteq \Bbb X$ has a point of total accumulation in $\Bbb X$. Replcaing "infinite" by merely "countably infinite" we get a characterisation of $\Bbb X$ being countably compact. So it's quite a natural notion IMHO, and better behaved than "condensation point"