Week 5 Compactness in metric and Euclidean spaces. Connectedness
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Metric spaces form a subclass of Hausdorff topological spaces. We can obtain further results about compact sets in this subclass. Recall the following from MATH21111 Metric spaces:
Definition: bounded set.
A subset \(A\) of a metric space \((X,d)\) is bounded if \(A\subseteq B_r(x)\) for some \(x\in X,\) \(r>0.\)
We reproduce a result from MATH21111, but with a new proof which refers to the Hausdorff property:
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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. □
Remark: it is not true that every closed and bounded subset of a metric space is compact. An easy counterexample is given by \(X\) endowed with a discrete metric, \(d(x,y) = \begin {cases} 0, & x=y, \\ 1, & x\ne y.\end {cases}\) The metric topology defined by \(d\) on \(X\) is the discrete topology which is not compact is \(X\) is infinite (exercise).
A more conceptual example of a non-compact closed and bounded set in a metric space is the closed unit ball of an infinite-dimensional Hilbert, or Banach, space. This will be discussed in the second part of MATH31010, Functional Analysis.
The next result gives us a highly non-trivial example of a compact set in a Euclidean space.
Theorem 5.2: the Heine-Borel Lemma.
The closed bounded interval \([0,1]\) is a compact subset of the Euclidean line \(\RR .\)
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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. □
Remark: it follows from the Heine-Borel Lemma that in every Euclidean space \(\RR ^n,\) a set is compact iff it is closed and bounded. This is because every such set is a subset of a cube of the form \([-M,M]^n\subseteq \RR ^n.\) The closed bounded cube can be shown to be compact by adapting the subdivision argument in Theorem 5.2.
We will, however, not extend the subdivision argument to \(n\) dimensions: in a short while, compactness of \([-M,M]^n\) will follow from the (baby) Tychonoff theorem which will say that a direct product of \(n\) compact spaces is compact.
Connectedness
So far, we have considered two topological properties: the Hausdorff property and compactness. The third topological property, and the final one that we will study in this course, is connectedness.
Definition: connected.
A topological space \(X\) is disconnected if
\(\exists U,V\) open in \(X:\) \(U\ne \emptyset ,\) \(V\ne \emptyset ,\) \(U\cap V=\emptyset ,\) \(U\cup V= X.\)
That is, a disconnected space is a disjoint union of two non-empty open sets.
The space \(X\) is connected if it is not disconnected.
Let us construct two disconnected spaces. We will obtain them as subspaces of the Euclidean line \(\RR :\)
Example.
Show that the subspace \(X=(0,1)\cup (2,3)\) of \(\RR \) is disconnected.
Solution: let \(U\) be the open interval \((0,1)\) and \(V\) be the open interval \((2,3).\) Then \(U\) and \(V\) are disjoint non-empty sets open in \(X\) whose union is \(X.\) Hence, by definition, \(X\) is disconnected.
Example.
Show that the subspace \(X=\{0,1\}\) of \(\RR \) is disconnected.
Solution. Consider the following non-empty disjoint subsets of \(X:\) \(U=\{0\}\) and \(V=\{1\}.\) We note that both sets are open in \(X.\) For example, \(\{0\} = (-\infty ,1)\cap X\) where \((-\infty , 1)\) is an open subset of \(\RR ,\) and so \(\{0\}\) is open in \(X\) by definition of subspace topology. Similarly, \(\{1\}= (0,+\infty )\cap X.\)
We have written \(X\) as a disjoint union of two non-empty open subsets, so by definition, \(X\) is disconnected.
Remark. By showing \(\{0\}\) and \(\{1\}\) to be open in \(\{0,1\},\) we have proved the following:
Claim.
The subspace \(\{0,1\}\) of the Euclidean line \(\RR \) has discrete topology. □
One can say that the discrete space \(\{0,1\}\) is the “canonical” example of a disconnected space. This point of view is supported by the following result.
Proposition 5.3: conditions equivalent to connectedness.
The following are equivalent for a topological space \(X:\)
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(i) \(X\) is connected.
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(ii) For all continuous functions \(f\colon X \to \RR ,\) the image \(f(X)\) is an interval.
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(iii) Every continuous function \(g\colon X \to \{0,1\}\) is constant.
Here \(\RR \) has Euclidean topology, and \(\{0,1\}\) has discrete topology.
Before proving the Proposition, we formally define “interval”.
Definition: interval.
An interval is a subset \(I\) of \(\RR \) such that
\(\forall x,y\in I,\) \(\forall t,\) \(x<t<y\) \(\implies \) \(t\in I.\)
Thus, together with any two points, an interval contains all intermediate points.
It is not difficult to establish the following classification of intervals:
Remark: the empty subset \(\emptyset \) of \(\RR \) can be written as \((a,a)\) and is an interval. Any singleton \(\{a\}\subseteq R\) can be written as \([a,a]\) and is an interval.
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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. □
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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). □
Here is an example where we use the Proposition 5.3 to establish that a space is connected.
Example: interval is connected.
Show that an interval \(I\subseteq \RR \) is connected.
Solution: we use condition (ii) from the Proposition. Let \(f\colon I \to \RR \) be a continuous function. To show that \(f(I)\) is an interval, take two points \(f(a)\) and \(f(b)\) of \(f(I),\) where \(a,b\in I.\) By the Intermediate Value Theorem from MATH11121 Mathematical Foundations and Analysis, all intermediate values between \(f(a)\) and \(f(b)\) belong to the image \(f(I).\) Thus, by definition of “interval”, \(f(I)\) is an interval. Hence by Proposition 5.3, \(I\) is connected, as claimed.
References for the week 5 notes
The bisection proof of the Heine-Borel Lemma that we present in Theorem 5.2, is given in [Armstrong, Theorem (3.3)] and [Sutherland, Exercise 13.15]. An alternative “creeping-along” proof can be found in [Armstrong, Theorem (3.3)] and [Sutherland, Exercise 13.9].
Our definition of “connected” is identical to [Willard, Definition 26.1]. We thus deviate slightly from [Sutherland] which uses condition (iii) of our Proposition 5.3 as a definition [Sutherland, Definition 12.1], and turns our definition into a theorem, [Sutherland, Proposition 12.3]. The way [Armstrong] defines “connected”, though equivalent to ours, requires the notion of closure of a set; but in our course closure comes after connectedness.
The definition of interval is elementary and is taken from [Smith, Definition 1.11]. The claim about classification of intervals is made in [Sutherland, Chapter 2: Notation and terminology].
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