Question: I am stumped on the following question

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> > PQRS is a parallelogram and ST=TR. What is the ratio of area of triangle QST to the area of parallelogram (Ans 1:4)
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![enter image description here](https://i.stack.imgur.com/o9JYN.png)

I need the height of the triangle, how would I get that? Any suggestions would be appreciated.

Answer: 
So you had two questions, one about the charts covering the space, and one about compactness.

The first has been answered, but i would like to add some details.
So you do see that the $U\_i = \phi\_i^{-1}(\mathbb{C})$ for $i=0,1$ cover $\mathbb{C} P^1$ right? My claim is now that we can restrict to the closed disks in $\mathbb{C}$. The cases $[0:1]$ and $[1:0]$ are simple and the only points outside $U\_{0,1} := U\_0 \cap U\_1$, so we restict to $[z:w] \in U\_{0,1}$. Now both $\phi\_i$ are defined at this point, and since $\phi\_0([z:w]) = \frac{1}{\phi\_1([z:w])}$ we can pick the one that has norm smaller or equal to one, so that the image lies in the unit disk. Since the point was arbitrary, every point has as $i$ such that its $\phi\_i$ image lies in the unit disk, q.e.d.

Regarding the second one, recall that the continuous image of a compact space is compact. Moreover it is well known that the sphere $S \subset \mathbb{C}^2$ is compact, and that $\mathbb{C}P^1$ is a quotient of this sphere. Summarizing, there is a surjective quotient map $\pi: S \rightarrow \mathbb{C}P^1 $, hence $\mathbb{C}P^1$ is the image of a compact space, hence compact.

The reason i add the second answer is that it is insightful and easily generalized to complex or real projective spaces of any dimension, even to Grassmanians (although that requires some adjustments) and i believe also you can replace $\mathbb{C}$ by any normed division algebra over $\mathbb{R}$.

Hope this helps.
Joachim
The argument you have outlined appeals to the following lemma of general topology.

**Lemma.** Let $X$ be a topological space. If there are (quasi)compact subsets $Y\_1, \ldots, Y\_n$ such that $Y\_1 \cup \cdots \cup Y\_n = X$, then $X$ is itself (quasi)compact.

*Proof.* Let $\{ U\_\alpha : \alpha \in I \}$ be an open cover of $X$. For each $i$, $\{ U\_\alpha \cap Y\_i \}$ is an open cover of $Y\_i$, so has a finite subcover $\{ U\_{\alpha\_j} \cap Y\_i : \alpha\_j \in J\_i \}$. Thus we have a finite subcover $\{ U\_{\alpha\_j} : 1 \le j \le n, \alpha\_j \in J\_i \}$ for $X$.