Let $ \overline{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ be the one-point compactification of $\mathbb{C}$. We know that in $\mathbb{C}$, every euclidean line is of the form
$$ \beta z + \overline{\beta} \overline{z} + \gamma = 0 $$ where $ \beta \in \mathbb{C}$ and $ \gamma \in \mathbb{R}$. Let us $L$ be the set of solutions of the equation above in $ \overline{\mathbb{C}} $. My question is that why $ \infty \in L $?
I know that we can construct a sequence $\{a_n\} \subset L \cap \mathbb{C}$ in the way that $ |a_i| $ s become greater and greater, and therefore become closer to $\infty$, and we can show that $\infty \in \overline{L}$. But from there on, I don't know how to prove $ \infty \in L $. I aimed to prove that $L$ is closed, but I didn't succeed. Also our professor remarked that $ \overline{\mathbb{C}}$ is compact. From the compactness of $ \overline{\mathbb{C}} $ I can just conclude that for any open cover $ \{ \mathcal{O}_i\} $ of $ L \cap \mathbb{C} $ there's a finite subcover $ \{ \mathcal{O}_{i_j}\}_{j=1}^n$ such that $ \infty \in \{ \mathcal{O}_{i_j}\}_{j=1}^n$. Also it was noted that $ L \cap \mathbb{C} $ is continuous, but I couldn't see how that might be helpful (also I wasn't sure what he meant by continuous, because I'd just seen continuity as an attribute of functions, and had not seen it for sets).