Show that mathbb C∪\∞\ (the extended complex plane…
Show that $\mathbb C\cup\{\infty\}$ (the extended complex plane $\overline{\mathbb C}$) is both an open and a closed set — in itself, as a topological space.
Explanation of Method
Neighborhoods of finite points in $\overline{\mathbb C}$ are the usual open disks; neighborhoods of $\infty$ are the sets $\{|z|>R\}\cup\{\infty\}$. Checking these two definitions directly shows every point of $\overline{\mathbb C}$ has a neighborhood inside $\overline{\mathbb C}$ (trivially — there is nowhere else to go), and its complement (the empty set) is open — giving both properties at once.
Step 1 · The topology on $\overline{\mathbb C}$
A set $U\subseteq\mathbb C$ is open in $\overline{\mathbb C}$ iff it is open in the usual sense; a set containing $\infty$ is open iff it has the form $\{|z|>R\}\cup\{\infty\}$ (or a union of such neighborhoods) for some $R>0$.
Step 2 · $\overline{\mathbb C}$ is open
For any $z_0\in\overline{\mathbb C}$: if $z_0\in\mathbb C$, the disk $B(z_0,1)\subseteq\overline{\mathbb C}$ trivially (every point is in $\overline{\mathbb C}$); if $z_0=\infty$, the neighborhood $\{|z|>1\}\cup\{\infty\}\subseteq\overline{\mathbb C}$, again trivially. So every point has a neighborhood inside $\overline{\mathbb C}$:
$$ \boxed{\overline{\mathbb C} \text{ is open}} $$
Step 3 · $\overline{\mathbb C}$ is closed
A set is closed iff its complement is open. The complement of $\overline{\mathbb C}$ (within $\overline{\mathbb C}$) is $\varnothing$, which is open (vacuously — see Problem 123). Hence:
$$ \boxed{\overline{\mathbb C} \text{ is closed}} $$
Step 4 · Equivalent argument via limit points
Every limit point of any subset of $\overline{\mathbb C}$ is, by definition, a point of $\overline{\mathbb C}$ itself (there is no "outside" to escape to). So $\overline{\mathbb C}$ trivially contains all its limit points, confirming closedness directly.
Remark · Clopen sets and connectedness
A set that is both open and closed is called clopen. In any topological space, the whole space and $\varnothing$ are always clopen. In $\overline{\mathbb C}$, these are in fact the only clopen sets, because $\overline{\mathbb C}$ is connected — it cannot be split into two disjoint nonempty open pieces.
Verification
Contrast with a proper subset: $\{\infty\}$ alone is closed in $\overline{\mathbb C}$ (its complement $\mathbb C$ is open) but not open (every neighborhood of $\infty$ necessarily contains infinitely many finite points too) — so $\{\infty\}$ is closed but not clopen, illustrating that the whole-space property in Step 2–3 is special to $\overline{\mathbb C}$ itself, not shared by its proper subsets.
Result
This is the general fact that the whole space is always clopen; in $\overline{\mathbb C}$'s case, connectedness guarantees it is also the only nontrivial clopen set alongside $\varnothing$.