10.10 Compactness[2CB]

The Heine-Borel Theorem [ 56 ] extends to this context.

(This theorem has a generalization in topological spaces, see 5).

E299

[0V4]Setting \(X=ℝ^ n\) and \(d\) the usual Euclidean distance, taken \(C⊆ ℝ^ n\), use the above theorem 299 to show (as a corollary) the usual Heine-Borel theorem [ 56 ] : \(C\) is compact if and only if it is closed and bounded.

Hidden solution: [UNACCESSIBLE UUID ’0V5’]

E299

[0V6] Show that if \(K⊂ X\) is compact then it is closed. Hidden solution: [UNACCESSIBLE UUID ’0V7’] (See 2 for the case of topological space)

E299

[0V8] Let \((X,d_ X)\) and \((Y,d_ Y)\) be metric spaces, with \((X,d_ X)\) compact; suppose that \(f:X→ Y \) is continuous and injective. Show that \(f\) is a homeomorphism between \(X\) and its image \(f(X)\).

Hidden solution: [UNACCESSIBLE UUID ’0V9’]

(See 4 for the case of topological space).

E299

[0VB] Let \(n≥ 1\) be natural. Let \((X_ i,d_ i)\) be compact metric spaces, for \(i=1,\ldots n\); choose \(y_{i,k}∈ X_ i\) for \(i=1,\ldots n\) and \(k∈ℕ\). Show that there exists a subsequence \(k_ h\) such that, for every fixed \(i\), \(y_{i,k_ h}\) converges, that is, the limit \(\lim _{h→∞} y_{i,k_ h}\) exists.

E299

[0VC] Difficulty:**.Let \((X_ i,d_ i)\) be compact metric spaces, for \(i∈ℕ\), and choose \(y_{i,k}∈ X_ i\) for \(i,k∈ℕ\). Show that there exists a subsequence \(k_ h\) such that, for every fixed \(i\), \(y_{i,k_ h}\) converges, that is, the limit \(\lim _{h→∞} y_{i,k_ h}\) exists.

E299

[0VD]Let be given a metric space \((X,d)\). As in 280 we define the disk \(D(x,\varepsilon ){\stackrel{.}{=}}\{ y∈ X, d(x,y)≤ \varepsilon \} \) (which is closed). \((X,d)\) is locally compact if for every \(x∈ X\) there exists \(\varepsilon {\gt}0\) such that \(D(x,\varepsilon )\) is compact. Consider this proposition.

«Proposition A locally compact metric space is complete. Proof Let \((x_ n)_ n⊂ X\) be a Cauchy sequence, then eventually its terms are distant at most \(\varepsilon \), so they are contained in a small compact disk, so there is a subsequence that converges, and then, by the result 9, the whole sequence converges. q.e.d. »

If you think the proposition is true, rewrite the proof rigorously. If you think it’s false, find a counterexample.

Hidden solution: [UNACCESSIBLE UUID ’0VF’]

E299

[0VG] Let \((X,d)\) be a metric space, and let \(C⊂ X\). Show that \(C\) is totally bounded if and only if \(\overline C\) is totally bounded. (See 299 for the definition of totally bounded). Hidden solution: [UNACCESSIBLE UUID ’0VH’]

E299

[2GB] Prerequisites:14.Let \((X,d)\) be a totally bounded metric space. Let \(E⊆ X\), then \(E\) is a metric space with the restricted distance \(\tilde d=d|_{E× E}\). Show that \((E,\tilde d)\) is totally bounded. (See 299 for the definition of totally bounded).

Hidden solution: [UNACCESSIBLE UUID ’2GC’]

E299

[0VJ] Prerequisites:2,9.Difficulty:*.

Let \((X,d)\) be a metric space such that every continuous function \(f:X→ℝ\) has maximum: show that the space is compact.

(See 3 for a rewording with \(X=ℝ^ n\).) [UNACCESSIBLE UUID ’0VK’]

Let \((X,d)\) be a metric space, and let \(A_ n⊆ X\) be compact non-empty subsets such that \(A_{n+1}⊆ A_ n\): then \(⋂_{n∈ℕ} A_ n≠ ∅\).

(This result can be derived from 3; but try to give a direct demonstration, using the characterization of ”compact” as ”sequentially compact ”, i.e. the first point in 299).

Hidden solution: [UNACCESSIBLE UUID ’0VQ’] [0VR] Let be given a metric space \((X,d)\) and its subset \(C⊆ X\) that is totally bounded, as defined in 299: show that \(C\) is bounded, i.e. for every \(x_ 0∈ C\) we have

or equivalently, for every \(x_ 0∈ C\) there exists \(r{\gt}0\) such that \(C⊆ B(x_ 0,r)\).

The opposite implication does not hold, as shown in 2 [0VS] Let \((X,d)\) be a metric space and let \(D⊆ X\), show that these clauses are equivalent:

• \(D\) is not totally bounded;

• there exists \(\varepsilon {\gt}0\) and there is a sequence \((x_ n)_ n⊆ D\) for which

  \[ ∀ n,m∈ℕ, ~ d(x_ n,x_ m)≥ \varepsilon \quad . \]

[0VT] Prerequisites:9.Let \(X=C^ 0([0,1])\) be the space of continuous and bounded functions \(f:[0,1]→ℝ\), endowed with the usual distance

We know that \((X,d_∞)\) is a complete metric space. Let

the disk of center \(0\) (the function identically zero) and radius 1. We know from 286 that it is closed, and therefore it is complete. Show that \(D\) is not totally bounded by finding a sequence \((f_ n)⊆ D\) as explained in 9.