# Correctness of reasoning about finiteness of degree of a covering map

Let $q$ be a covering $q \colon \mathbb{R} P^{2n} \to X$, where $X$ is path-connected. Call $V_x$ the open nbhd of $x \in X$ given by the definition of covering map.

We first note that $X$ must be compact, being the image through a continuous map of a compact set. We prove now that the degree of $q$ must be finite:

Suppose the contrary: so we have that for every point of $X$ there exists a neighbourhood $V_x$ such that its preimage is an infinite disjoint union of open sets in $\mathbb{R} P^{2n}$. The union $\bigcup_{x \in X}V_x$ is an open cover of $X$, so there exists a finite subcover, call it $\bigcup_{j=1}^n V_{x_j}$.

Wlog we can assume that the cover is minimal, i.e. there are not elements of the subcover entirely covered by other elements of the subcover ($\nexists i \in \{1,..,n\}$ s.t. $V_{x_i} \subset \bigcup_{j=1,i\neq j}^nV_{x_j}$).

The preimage $$\left\lbrace \bigcup_{j=1}^n\left( \bigcup_{\tilde{x} \in q^{-1}(\{ x_j \})} \tilde{V}_{ \tilde{x} }\right) \right\rbrace$$ is an open covering of $\mathbb{R} P^{2n}$, where we denoted with $\tilde{V}_{\tilde{x}}$ one of the disjoint open neighbourhood of the pre image of one of the $V_{x_j}$.

By the compactness of $\mathbb{R}P^{2n}$ it must admit a finite subcover, hence being the first union finite, only a finite quantity of elements in the second union can belong to the sub cover, because they are infinite many.

In particular there must exist $\tilde{V}_{\tilde{x}}$ s.t. it is not in the subcover, and so it must be contained in the union of some other elements of the subcover. Considering the image of $\tilde{V}_{\tilde{x}}$ and of the union we get an absurd because we assumed our cover to be minimal.

The Wlog part for me is reasonable, because I have a finite number of open to check and I can eliminate redundant elements. In this way I should get my absurd. Is this correct?