Can a finite set E⊂mathbb C be open?

Can a finite set $E\subset\mathbb C$ be open?

Step 1 · Suppose $E$ is finite and open

Suppose, for contradiction, that $E=\{z_1,\ldots,z_n\}$ is a nonempty finite open set.

Step 2 · Openness forces a disk around each point

Since $E$ is open, for $z_1\in E$ there is $\varepsilon>0$ with $\ B(z_1,\varepsilon)=\{z:|z-z_1|<\varepsilon\}\subseteq E$.

Step 3 · Every disk is infinite

The disk $B(z_1,\varepsilon)$ contains, for instance, every point $z_1+re^{i\theta}$ with $0\le r<\varepsilon$, $\theta\in[0,2\pi)$ — an uncountably infinite set of points.

Step 4 · Contradiction

So $B(z_1,\varepsilon)\subseteq E$ forces $E$ to contain infinitely many points, contradicting that $E$ is finite.

$$ \boxed{\text{No nonempty finite set is open (in the standard topology on } \mathbb C\text{)}} $$

The empty set is the sole exception

$\varnothing$ is open vacuously: the defining condition ("for every $z\in\varnothing$, there is $\varepsilon>0$ with $B(z,\varepsilon)\subseteq\varnothing$") has no points to check, so it holds automatically.

What finite sets actually are: closed, not open

Every finite set $E=\{z_1,\ldots,z_n\}$ is closed: it has no limit points at all (any point outside $E$ has a disk avoiding $E$, since $E$ is a finite set of isolated points), so it vacuously/trivially contains all its limit points. Combined with Step 4, a nonempty finite set is closed but never open.

Verification

Example: $E=\{0,1,i\}$. It is closed and bounded (hence compact, by Heine–Borel), but not open — no disk around $0$, however small, avoids the other two points $1,i$ while still fitting entirely inside the three-point set $E$, since $B(0,\varepsilon)$ always contains points other than $0,1,i$.