Compactness & Contraction Mappings

Compactness and completeness solve different problems. Completeness prevents a Cauchy approximation from falling out of the space. Compactness guarantees that every infinite sequence has a convergent subsequence and therefore turns many limiting arguments into finite ones. The contraction theorem needs completeness, not compactness; compactness enters other existence arguments.

Equivalent compactness tests in metric spaces #

For a subset $K$ of a metric space, the following are equivalent:

• every open cover of $K$ has a finite subcover;
• every sequence in $K$ has a subsequence converging to a point of $K$;
• $K$ is complete and totally bounded.

Total boundedness means that for every $\varepsilon>0$, finitely many $\varepsilon$-balls cover $K$. It is stronger than ordinary boundedness. The distinction is invisible in finite-dimensional Euclidean space, where a set is compact exactly when it is closed and bounded.

In an infinite-dimensional normed space, the closed unit ball is not compact. Riesz's lemma constructs unit vectors separated by a fixed positive distance, so the sequence has no Cauchy subsequence. This is why function-space compactness theorems need extra regularity or tail control.

What compactness buys #

If $K$ is compact and $f:K\to\mathbb R$ is continuous, then $f(K)$ is compact, so $f$ is bounded and attains its minimum and maximum. If a continuous map is bijective from compact $K$ into a Hausdorff space, its inverse is automatically continuous. A continuous function on compact $K$ is uniformly continuous: one input tolerance works over the whole set.

These conclusions fail when compactness is replaced by boundedness alone. For example, $f(x)=1/x$ is continuous but unbounded on the bounded set $(0,1)$; the missing endpoint is exactly where compactness fails.

Banach's contraction principle #

Let $(X,d)$ be complete and let $T:X\to X$ satisfy

$$d(Tx,Ty)\le qd(x,y),\qquad 0\le q<1.$$

Starting from any $x_0$, define $x_{n+1}=Tx_n$. Repeated contraction gives

$$d(x_{n+1},x_n)\le q^n d(x_1,x_0).$$

Summing the geometric tail shows that $(x_n)$ is Cauchy:

$$d(x_m,x_n)\le \frac{q^n}{1-q}d(x_1,x_0),\qquad m>n.$$

Completeness provides a limit $x_*$. Continuity of $T$ gives $Tx_*=x_*$. If $y_*$ is another fixed point, then $d(x_*,y_*)\le qd(x_*,y_*)$, forcing equality to zero. The fixed point is unique.

Two practical error bounds follow:

$$d(x_n,x_*)\le\frac{q^n}{1-q}d(x_1,x_0)$$

uses the first step, while

$$d(x_n,x_*)\le\frac{q}{1-q}d(x_n,x_{n-1})$$

uses the most recent step and can be monitored during computation.

Worked example: solving $x=\cos x$ #

On $X=[0,1]$ with the usual metric, $T(x)=\cos x$ maps $X$ into itself and $|T'(x)|=|\sin x|\le\sin1<1$. Thus $T$ is a contraction with $q=\sin1\approx0.8415$. Starting at $x_0=1$ gives

$$1, 0.5403, 0.8576, 0.6543, 0.7935,ldots$$

and converges to the unique fixed point $x_*\approx0.739085$. The alternating appearance does not threaten convergence; the contraction controls distance, not monotonicity.

How fixed-point proofs fail #

• If $q=1$, nonexpansiveness alone gives no uniqueness and may give no convergence.
• If the proposed domain is not invariant, an iterate can leave the region where the derivative bound holds.
• If the space is incomplete, the Cauchy iterates may converge only in its completion.
• A local derivative bound proves only a local contraction unless the iterates are shown to stay inside the neighborhood.
• Floating-point termination should use a derived residual/error criterion, not merely a fixed iteration count.