Compact Sets

Compact

\iff

closed + bounded (Heine–Borel):

[0,1]

is covered by the first 6 intervals of its open cover, while

(0,1]

fails at the gap near 0 and

[0,\infty)

fails at the infinite tail. Sequential compactness via Bolzano–Weierstrass is proved in full; the open-cover definition's equivalence is shown at the idea level.

Examples & Exercises

Show that $[0,1]$ is sequentially compact: every sequence in $[0,1]$ has a convergent subsequence whose limit stays in $[0,1]$ (Bolzano–Weierstrass).

(a_n) \subset [0,1] \Rightarrow \exists\,(a_{n_k}) \to L \in [0,1]

K \text{ compact } \iff K \text{ closed and bounded} \quad \text{(Heine--Borel)}

k = 6

: the union

[-0.18, 1.18]

of the first 6 open intervals covers

[0,1]

— a finite subcover exists.

Show proof

Let $(a_n)$ be any sequence in $[0,1]$ . It is bounded, so the Bolzano–Weierstrass theorem gives a convergent subsequence $(a_{n_k}) \to L$ .
Since $0 \le a_{n_k} \le 1$ for every $k$ , passing to the limit preserves the inequalities, giving $0 \le L \le 1$ . Hence $L \in [0,1]$ — a closed interval never loses a limit point.
Every sequence in $[0,1]$ therefore has a convergent subsequence with its limit in $[0,1]$ : $[0,1]$ is sequentially compact.
In $\mathbb{R}$ , sequential compactness ⟺ compactness (Heine–Borel: closed + bounded). The open-cover picture tells the same story: the union $[-0.18, 1.18]$ of the first 6 intervals $U_1, \dots, U_6$ already covers all of $[0,1]$ — the whole cover is never needed. (The proof that the open-cover definition is equivalent is shown at the idea level only.)

Show that $(0,1]$ is not compact: the sequence $a_n = 1/n$ has no subsequence converging to a point of $(0,1]$ .

a_n = \frac{1}{n} \to 0, \qquad 0 \notin (0,1]

\bigcup_{k=1}^{N} \left(\tfrac{1}{k}, 2\right) = \left(\tfrac{1}{N}, 2\right): \text{ the gap } \left(0, \tfrac{1}{N}\right] \text{ is never covered}

k = 8

: even all 8 intervals leave

(0, 1/8]

uncovered — with

0

missing, no finite subcover exists.

Show proof

Put $a_n = 1/n$ . Then $a_n \in (0,1]$ for every $n$ , and $a_n \to 0$ .
A subsequence of a convergent sequence converges to the same limit, so every subsequence $(a_{n_k})$ of $(a_n)$ also converges to $0$ .
But $0 \notin (0,1]$ , so no subsequence converges to a point of $(0,1]$ . Hence $(0,1]$ is not sequentially compact, and not compact.
The open-cover picture shows the same failure: however many intervals we take from the cover — the first $N$ of $U_k = \left(\frac{1}{k}, 2\right)$ ( $k = 1, \dots, 8$ ) — their union is always of the form $\left(\frac{1}{N}, 2\right)$ , leaving the gap $\left(0, \frac{1}{N}\right]$ — at $k = 8$ that gap is $(0, 1/8]$ . Closing the gap would require the point $0$ , which is not in the set.

Prove that if $K$ is compact and $f$ is continuous on $K$ , then the image $f(K)$ is compact.

y_n = f(x_n), \quad x_{n_k} \to x \in K \Rightarrow f(x_{n_k}) \to f(x) \in f(K)

K = [0,1] \text{ compact} \Rightarrow f(K) \text{ compact } \Rightarrow \max f, \min f \text{ exist (EVT)}

The engine of the Extreme Value Theorem:

[0,1]

is compact, so the continuous image

f([0,1])

is compact — max and min exist.

Show proof

Let $y_n \in f(K)$ be any sequence, and pick $x_n \in K$ with $y_n = f(x_n)$ for each $n$ .
Since $K$ is compact, $(x_n)$ has a subsequence $(x_{n_k})$ converging to some $x \in K$ (sequential compactness).
Continuity of $f$ at $x$ gives $f(x_{n_k}) \to f(x)$ . So the subsequence $y_{n_k} = f(x_{n_k})$ converges to $f(x) \in f(K)$ .
Every sequence in $f(K)$ therefore has a subsequence converging to a point of $f(K)$ : $f(K)$ is sequentially compact, hence compact. This is the engine of the Extreme Value Theorem (the evt scene) — for continuous $f$ on $[0,1]$ , the set $f([0,1])$ is compact, hence closed and bounded in $\mathbb{R}$ , so its supremum and infimum are actually attained.

Part of these learning paths

Examples & Exercises