Regard mathbb C as a subset of the extended complex plane…

Regard $\mathbb C$ as a subset of the extended complex plane $\overline{\mathbb C}=\mathbb C\cup\{\infty\}$. Is $\mathbb C$ closed?

Step 1 · $\mathbb C$ is open in $\overline{\mathbb C}$

For every $z_0\in\mathbb C$, the disk $B(z_0,1)\subseteq\mathbb C$, so $\mathbb C$ is open — this part is uncontroversial.

Step 2 · Test closedness via the complement

$\mathbb C$ is closed in $\overline{\mathbb C}$ iff $\overline{\mathbb C}\setminus\mathbb C=\{\infty\}$ is open in $\overline{\mathbb C}$.

Step 3 · $\{\infty\}$ is not open

Every neighborhood of $\infty$ has the form $\{|z|>R\}\cup\{\infty\}$ for some $R>0$ — and this set always contains infinitely many finite points, never just $\{\infty\}$ alone. So no neighborhood of $\infty$ is contained in $\{\infty\}$:

$$ \{\infty\} \text{ is not open} \quad\Longrightarrow\quad \boxed{\mathbb C \text{ is not closed in } \overline{\mathbb C}} $$

Step 4 · Confirm via a limit point outside $\mathbb C$

Take $z_n=n$ for $n=1,2,3,\ldots$ — a sequence entirely within $\mathbb C$. In the topology of $\overline{\mathbb C}$ (where neighborhoods of $\infty$ are exactly the sets $\{|z|>R\}\cup\{\infty\}$), this sequence converges:

$$ \lim_{n\to\infty}z_n=\infty $$

Since $\infty\notin\mathbb C$, the set $\mathbb C$ fails to contain one of its own limit points — confirming it is not closed.

Summary of status in $\overline{\mathbb C}$

General principle

A subset $E\subseteq\mathbb C$ is closed in $\overline{\mathbb C}$ if and only if $E$ is compact in $\mathbb C$ (closed and bounded, by Heine–Borel). Since $\mathbb C$ itself is unbounded, it fails compactness and is therefore not closed in $\overline{\mathbb C}$ — even though it is closed (trivially) in itself.