Reading through Big Rudin, I have come across the following statement in the proof of Theorem 1.40:

Let $S \subset \mathbb{C}$ be a closed set [in the topology induced by the complex modulus]. Let $\Delta = \{z \in \mathbb{C}: |z-\alpha|\leq \epsilon\}$ be a closed disc about the point $z \in \mathbb{C}-S$. Then $\mathbb{C}-S$ is given by a countable union of closed discs $\Delta$.

An (arbitrary) open set is the union of a countable number of closed sets? This seems like an error to me, but I can find no mention of it online. For what its worth, the proof seems to work if we take $\Delta$ to be an open disc, as the standard topology on $\mathbb{C}$ is second-countable.

Edit: Does Anyone know of a good errata for the book? The only ones making themselves known are for Baby Rudin.


An arbitrary open set of $\mathbb{C}$ is a countable union of open balls - specifically, if $U\subseteq \mathbb{C}$ is open, then $U$ is the union of the "rational" open balls contained in $U$, that is, those open balls whose centers are complex numbers whose real and imaginary parts are rational, and whose radii are rational (since $\mathbb{Q}$ is countable, there can only be countably many such open balls). Now note that any open ball is a countable union of closed balls - specifically, $\{z\in\mathbb{C}:|z-\alpha|<\epsilon\}$ is equal to the union $$\bigcup_{n\in\mathbb{N}}\{z\in\mathbb{C}:|z-\alpha|\leq\epsilon-\tfrac{1}{n}\}.$$ Lastly, note that a countable union of countable unions is still a countable union.

  • 2
    $\begingroup$ Would you happen to know Rudin's reason for choosing $\Delta$ to be a closed disc rather than an open disc? I, too, observed that taking $\Delta$ to be an open disc works fine. $\endgroup$ – user193319 Dec 15 '17 at 18:47

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