Proof. Let \(K\) be a compact subset of a metric space \(X.\) The metric topology on \(X\) is Hausdorff, Proposition 3.2, and compacts are closed in Hausdorff spaces, Proposition 4.4, so \(K\) is closed in \(X.\)

To show that \(K\) is bounded, fix any point \(x\) of \(X\) and consider the collection \(\mathscr C = \{B_r(x)\}_{r\in \RR _{>0}}\) of open balls. Clearly, \(\bigcup \mathscr C=X\) and so \(\mathscr C\) covers \(K.\) By Criterion of compactness 4.1, there exists a finite subcollection \(\{B_{r_1}(x),\dots ,B_{r_n}(x)\}\) of \(\mathscr C\) which still covers \(K:\) that is,

\[ K \subseteq B_{r_1}(x)\cup \dots \cup B_{r_n}(x) = B_R(x) \]

where \(R = \max (r_1,\dots ,r_n).\) We have shown that \(K\) is a subset of an open ball, that is, \(K\) is bounded. □

Proof. There are several standard proofs of this result; we will present the proof by bisection, or halving the interval. Alternative proofs can be found in the literature.

Assume for contradiction that there exists a collection \(\mathscr C\) of open subsets of \(\RR \) such that \([0,1]\) is covered by \(\mathscr C,\) yet is not covered by any finite subcollection of \(\mathscr C.\) Then at least one of the two halves, the closed subintervals

\([0,\frac 12]\) and \([\frac 12,1]\)

of \([0,1],\) has no finite subcover in \(\mathscr C:\) indeed, if \([0,\frac 12]\) were covered by finite \(\mathscr C_1\subseteq \mathscr C,\) and \([\frac 12,1],\) by finite \(\mathscr C_2\subseteq \mathscr C,\) then the whole of \([0,1]\) would be covered by \(\mathscr C_1\cup \mathscr C_2\) which is a finite subcollection of \(\mathscr C.\) We therefore let

\([a_1, b_1],\) where \(0\le a_1\le b_1 \le 1\) and \(b_1-a_1=\frac 12,\)

denote one of the halves of \([0,1]\) which is not covered by any finite subcollection of \(\mathscr C.\) We can now apply the same argument to the closed bounded interval \([a_1,b_1]\) and obtain

\([a_2,b_2],\) where \(a_1\le a_2\le b_2 \le b_1\) and \(b_2-a_2=\frac 1{2^2},\)

one of the halves of \([a_1,b_1]\) which is not covered by any finite subcollection of \(\mathscr C.\) Continuing this process, we will construct, for all \(n\ge 1,\) the interval

\([a_n,b_n],\) where \(a_{n-1}\le a_n\le b_n \le b_{n-1}\) and \(b_n-a_n=\frac 1{2^n},\)

which is not covered by any finite subcollection of \(\mathscr C.\)

Observe that the sequence \(a_1\le a_2\le a_3\le \dots \) is increasing and bounded, as all its terms lie in \([0,1].\) By a result from Year 1 Foundations of Mathematics course, such a sequence converges to a limit \(\ell ,\) and moreover \(\ell \in [0,1],\) \(a_n\le \ell \) for all \(n.\) Note also that \(\lim _{n\to \infty } b_n = \lim _{n\to \infty } a_n+\frac 1{2^n} = \ell \) and \(\ell \le b_n\) for all \(n.\)

The point \(\ell \) of \([0,1]\) must be covered by some set \(U\in \mathscr C.\) Since \(U\) is open, \(U\supseteq (\ell -\varepsilon ,\ell +\varepsilon )\) for some \(\varepsilon >0.\) Take \(n\) such that \(\frac 1{2^n}<\varepsilon .\) Then \(\ell -\varepsilon < \ell - \frac 1{2^n} \le a_n \le \ell \) and so \(a_n\in (\ell -\varepsilon ,\ell +\varepsilon ).\) Similarly, \(b_n\in (\ell -\varepsilon ,\ell +\varepsilon ),\) see Figure 5.1 for illustration.

Thus, the interval \([a_n,b_n]\) has a finite subcover in \(\mathscr C\) — in fact, a cover by just one set, \(U\in \mathscr C\) — contradicting the construction of \([a_n,b_n].\) This contradiction proves the Theorem. □

Sketch of proof of the Claim (not given in class). It is clear that every set \((a,b),\) \([a,b),\dots ,\) \(\RR ,\) listed in the claim, is an interval. Conversely, assume that \(I\) is an interval in \(\RR \) and let \(a=\inf I,\) \(b=\sup I.\) If both \(a\) and \(b\) are real numbers and not \(\pm \infty \) and so \(I\) is non-empty and bounded, one uses the definition of the “least upper bound” (\(\sup \)) and the “greatest lower bound” (\(\inf \)) to show that \((a,b)\subseteq I\subseteq [a,b],\) which leaves four possibilities for \(I.\) The remaining cases when one or both of \(a,b\) is infinite are handled similarly. Details are left to the student. □

Proof of Proposition 5.3. Recall for use in the proof that the preimage of intersection is the intersection of preimages; same for the union and the complement.

(i) \(\Rightarrow \) (ii): to prove the contrapositive, we must assume that there exists a continuous function \(f\colon X \to \RR \) such that \(f(X)\) is not an interval. This means that there are \(a,b\in f(X)\) and a real number \(t\) such that \(a<t<b\) and \(t\notin f(X).\) Consider

\[ U = f^{-1}((-\infty ,t)), \quad V = f^{-1}((t,+\infty )). \]

The set \(U\subseteq X\) is open in \(X\) because \(U\) is the preimage of the set \((-\infty , t),\) open in \(\RR ,\) under a continuous function \(f.\) Furthermore, \(U\) is not empty, as \(f(U)\ni a.\) Similarly, \(V\) is open in \(X\) and not empty; yet \(U\cap V\) is

\[ f^{-1}((-\infty , t))\cap f^{-1} ((t,+\infty )) = f^{-1}((-\infty , t)\cap (t,+\infty )) = f^{-1}(\emptyset ), \]

so \(U\cap V=\emptyset \) and \(U,V\) are disjoint. Finally,

\[ U\cup V = f^{-1}(\RR \setminus \{t\}) = X \setminus f^{-1}(\{t\}), \]

yet \(t\notin f(X)\) by assumption, so \(f^{-1}(t)=\emptyset \) and \(U\cup V=X.\) The construction of \(U,V\) shows that \(X\) is disconnected by definition. We have shown not(ii) \(\Rightarrow \) not(i).

(ii) \(\Rightarrow \) (iii): let \(g\colon X \to \{0,1\}\) be continuous. As in an earlier example, we can view the discrete set \(\{0,1\}\) as a subspace of the Euclidean line \(\RR .\) The inclusion map \(\mathrm {in}\colon \{0,1\}\to \RR \) is continuous by Proposition 2.7, and so we have a continuous map \(X \xrightarrow {g} \{0,1\} \xrightarrow {\mathrm {in}} \RR .\) By (ii), the image of this map must be an interval in \(\RR ;\) yet this image is a subset of \(\{0,1\},\) and such an interval can be at most one point. Thus, \(g\) must be constant, proving (iii).

(iii) \(\Rightarrow \) (i): to prove the contrapositive, assume that \(X\) is disconnected, so that \(X=U\cup V\) where \(U,\) \(V\) are disjoint non-empty open sets. This allows us to define the following function (as illustrated in Figure 5.2):

\[ g\colon X \to \{0,1\}, \quad g(x) = \begin {cases} 0, & x\in U, \\ 1, & x\in V. \end {cases} \]

We check that \(g\) is continuous by calculating the preimage of every open subset of \(\{0,1\}.\) There are exactly \(4\) open subsets of the discrete space \(\{0,1\},\) and we have \(g^{-1}(\emptyset )=\emptyset ,\) \(g^{-1}(\{0\})=U,\) \(g^{-1}(\{1\})=V\) and \(g^{-1}(\{0,1\})=X.\) In each case, the preimage is an open subset of \(X,\) so \(g\) is continuous by definition. We have constructed a non-constant continuous function \(X\to \{0,1\},\) proving not(i) \(\Rightarrow \) not(iii). □