Why the extended complex plane $\mathbb{C_\infty} = \mathbb{C}\cup \{\infty\}$ is compact while $\mathbb{C}$ is not?

Question

I think this is due to the fact the extended complex plane can be stereographically identified with the unit sphere in $\mathbb{R^3}$.

However, an exact and proper explanation is required.

Answer 1

This is called the one-point compactification and it is used to make a non-compact topological space, compact. Is is done in the following manner:

First, let $X$ be a noncompact topological space, and $\infty \notin X$, we construct a the space $X'=X\cup \{\infty\}$ and endow it with the cofinite topology, i.e. the topology is $$\tau = \tau_X\cup \{U\cup \{\infty \}: U \text{ is open in $X$ and }|X\setminus U|<\omega \}$$
Where $\tau_X$ is the original topology on $X$, clearly $\tau \supset \tau_X$, nevertheless consider and open cover $\mathcal{U}$, there exist some $U\in \mathcal{U}$ such that $\infty \in U$, then $X\setminus U$ must be finite, For each $x\in X\setminus U$ choose one $V_x\in \mathcal{U}$ such that $x\in V_x$ consider now the collection of sets $\{V_x : x\in X\setminus U \}\cup \{ U\}$ is a finite open cover of $X'$. Thus we conclude $X'$ is compact. This is the general case, in your particular question $\mathbb{C}_\infty=X'$ and $\mathbb{C}=X$

Answer 2

Topologically, $\{B(0,R) : R>0\}$ is an open cover of $\mathbb{C}$ which does not have a finite subcover, but it is not an open cover of $\mathbb{C}_\infty$.