According to the definition on Rudin' Principles of Mathematicial Analysis, closed set is defined as:
$E$ is closed if every limit point of $E$ is a point of $E$.
Then I have a question: if a set has no limit point, is it necessarily closed? I think this idea doesn't contradict the definition of closed sets. Am I right?
And another question is #10 in Chap.2 of Rudin's:
Let $X$ be an infinite set. For $p\in X$ and $q\in X$, define
$d(p,q)=0$ when $p=q$. Otherwises,$d(p,q)=1$.
I have known that every set in the metric space is closed. To prove it, we can show that the complement of any set is open, which is quite trivial. My question is: how can we prove that every set is closed just by checking according to the definition of closed set, i.e. how can we show that every limit point of any arbitrary set $E$ in the metric space is a point of $E$?
Thanks in advanced!
Best Answer
You are absolutely correct that a set $E$ without limit points is closed, vacuously, since the empty set of limit points of $E$ is necessarily a subset of $E$. In fact, this gives us an alternate way to show that every set in a discrete space (which is the sort of space $X$ you describe) is closed. Take any subset $E$ of $X$ and any point $x\in X$. The open $d$-ball centered at $x$ of radius $\frac12$ is simply $\{x\},$ which contains at most one point of $E$, so $x$ is not a limit point of $E$. Since $x$ was arbitrary, then $E$ is closed.