Characterization of Compactness in R
Proof
\(\Rightarrow\): Assume a set \(K\) is compact. We’ll prove that \(K\) is closed and bounded. Suppose for the sake of contradiction that \(K\) is not a bounded set. The definition of a compact set states that every sequence \((a_n)\) in \(K\) has a subsequence \(a_{n_k}\) that converges to a limit that is also in \(K\). So the idea is to derive a contradiction from finding some sequence that grows to infinity in a way that it cannot have a convergent subsequence. Since \(K\) is not a bounded set, then there must exists an element \(x_1 \in K\) such that \(|x_1| > 1\). Similarly, there must exists some element \(x_2 \in K\) with \(x_2 > 2\). In general, given any \(n \in \mathbf{N}\), we can produce \(x_n \in K\) such that \(|x_n| > n\). \(x_1, x_2, ... x_n, ... = (x_n)\) is a sequence in \(K\). Since \(K\) is compact, then there must exist a convergent subsequence \((x_{n_k})\). But all the elements in \((x_{n_k})\) must satisfy \(|x_{n_k}| > n_k\) (why? see this) and consequently \((x_{n_k})\) is unbounded. But theorem 2.3.2 states that convergent sequences must bounded. Therefore, we’ve arrived at a contradiction and \(K\) must be a bounded set.
To show that \(K\) is closed, we must show that it includes all of its limit points. Suppose \(x\) is a limit point of \(K\) such that \((x_n) \rightarrow x\) and \((x_n)\) is contained in \(K\). We will show that \(x \in K\). Since \(K\) is compact, then \((x_n)\) must have a convergent subsequence \((x_{n_k})\) with its limit is in \(K\). But by theorem 2.5.2, subsequences of a convergent sequence must converge to the same limit and so \((x_n)\) must converge to the same limit in \(K\). Therefore \(K\) is closed as we wanted to show.
\(\Leftarrow\): Assume \(K\) is closed and bounded. We will show that \(K\) is compact. Let \((x_n)\) be a sequence in \(K\). We will show that there exists a subsequence that converges to a limit also in \(K\). Since \(K\) is bounded, then there exists an \(M > 0\) such that \(\lvert k\rvert \leq M\) for all \(k \in K\). This also implies that \(\lvert x \rvert \leq M\) for all \(x \in (x_n)\). This means that the sequence \((x_n)\) is bounded. By the Bolzano-Weierstrass theorem, then \((x_n)\) must contain a convergent subsequence \((x_{n_k})\) where \((x_{n_k}) \rightarrow l\). Moreover, we also know that \(K\) is closed. This means that \(K\) contains all of its limit points and therefore, \(l \in K\). From this we can conclude that \(K\) is compact as required. \(\blacksquare\)
Notes
The first part of the proof is from the book. The second part is Exercise 3.3.3 and it’s my own and could contain mistakes!
Other Definitions and Properties
- Open Sets
- Limit Points
- Closed Sets
- Closed Sets
- Closure
- Absolute Value Function
- Sequences, Subsequences and Convergence
- Series and Series Convergence
- Show the Limit Template
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