If S1 and S2 are are compact sets. Why is S1 union S2 also compact?

To determine why the union of two compact sets, S1 and S2, is also compact, we need to use the definition of compactness.

A set is compact if it is closed and bounded.

Let's break down the proof into two parts:

1. Show that S1 union S2 is closed:
To prove that S1 union S2 is closed, we need to show that its complement, (S1 union S2)' (the set of all points not in S1 union S2), is open.

Since S1 and S2 are both closed sets, their complements, S1' and S2', respectively, are open.

Therefore, the complement of S1 union S2, (S1 union S2)', can be expressed as the intersection of S1' and S2':

(S1 union S2)' = (S1') intersect (S2')

The intersection of two open sets is always open. Therefore, (S1 union S2)' is open, which implies that S1 union S2 is closed.

2. Show that S1 union S2 is bounded:
To prove that S1 union S2 is bounded, we need to show that there exists some ball B(y, r) that contains S1 union S2.

Since S1 and S2 are compact, they are both bounded. Let's denote their respective bounding balls as B(x1, r1) and B(x2, r2).

Now, consider the union of these bounding balls, B(x1, r1) union B(x2, r2). This union will include all points in S1 union S2 since S1 and S2 are contained within their respective balls.

Since a union of balls is a ball itself, we can define a new ball B(y, r), where y is the center of B(x1, r1) union B(x2, r2), and r is the maximum of r1 and r2.

Therefore, S1 union S2 is bounded.

Combining the results from both parts, we have shown that S1 union S2 is closed and bounded. Hence, S1 union S2 is also compact.