Week 5 notes

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:

Proposition 5.1.

In a metric space, a compact set is closed and bounded.

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 $C = {B_{r} (x)}_{r \in R_{> 0}}$ of open balls. Clearly, $⋃ C = X$ and so $C$ covers $K .$ By Criterion of compactness 4.1, there exists a finite subcollection ${B_{r_{1}} (x), \dots, B_{r_{n}} (x)}$ of $C$ 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, $d (x, y) = {\begin{cases} 0, & x = y, \\ 1, & x \neq y . \end{cases}$ 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 $R .$

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 $C$ of open subsets of $R$ such that $[0, 1]$ is covered by $C,$ yet is not covered by any finite subcollection of $C .$ Then at least one of the two halves, the closed subintervals

$[0, \frac{1}{2}]$ and $[\frac{1}{2}, 1]$

of $[0, 1],$ has no finite subcover in $C :$ indeed, if $[0, \frac{1}{2}]$ were covered by finite $C_{1} \subseteq C,$ and $[\frac{1}{2}, 1],$ by finite $C_{2} \subseteq C,$ then the whole of $[0, 1]$ would be covered by $C_{1} \cup C_{2}$ which is a finite subcollection of $C .$ We therefore let

$[a_{1}, b_{1}],$ where $0 \leq a_{1} \leq b_{1} \leq 1$ and $b_{1} - a_{1} = \frac{1}{2},$

denote one of the halves of $[0, 1]$ which is not covered by any finite subcollection of $C .$ We can now apply the same argument to the closed bounded interval $[a_{1}, b_{1}]$ and obtain

$[a_{2}, b_{2}],$ where $a_{1} \leq a_{2} \leq b_{2} \leq b_{1}$ 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 $C .$ Continuing this process, we will construct, for all $n \geq 1,$ the interval

$[a_{n}, b_{n}],$ where $a_{n - 1} \leq a_{n} \leq b_{n} \leq b_{n - 1}$ and $b_{n} - a_{n} = \frac{1}{2^{n}},$

which is not covered by any finite subcollection of $C .$

Observe that the sequence $a_{1} \leq a_{2} \leq a_{3} \leq \dots$ 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 $ℓ,$ and moreover $ℓ \in [0, 1],$ $a_{n} \leq ℓ$ for all $n .$ Note also that $lim_{n \to \infty} b_{n} = lim_{n \to \infty} a_{n} + \frac{1}{2^{n}} = ℓ$ and $ℓ \leq b_{n}$ for all $n .$

The point $ℓ$ of $[0, 1]$ must be covered by some set $U \in C .$ Since $U$ is open, $U \supseteq (ℓ - 𝜀, ℓ + 𝜀)$ for some $𝜀 > 0.$ Take $n$ such that $\frac{1}{2^{n}} < 𝜀 .$ Then $ℓ - 𝜀 < ℓ - \frac{1}{2^{n}} \leq a_{n} \leq ℓ$ and so $a_{n} \in (ℓ - 𝜀, ℓ + 𝜀) .$ Similarly, $b_{n} \in (ℓ - 𝜀, ℓ + 𝜀),$ see Figure 5.1 for illustration.

Figure 5.1: the contradiction arrived at in the proof of the Heine-Borel Lemma

Thus, the interval $[a_{n}, b_{n}]$ has a finite subcover in $C$ — in fact, a cover by just one set, $U \in C$ — 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 $R^{n},$ 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 $[- M, M]^{n} \subseteq R^{n} .$ 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 $X :$ $U \neq \emptyset,$ $V \neq \emptyset,$ $U \cap V = \emptyset,$ $U \cup V = X .$

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 $R :$

Example.

Show that the subspace $X = (0, 1) \cup (2, 3)$ of $R$ 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 $R$ 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 $R,$ 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 $R$ 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 $X :$

(i) $X$ is connected.
(ii) For all continuous functions $f : X \to R,$ the image $f (X)$ is an interval.
(iii) Every continuous function $g : X \to {0, 1}$ is constant.

Here $R$ 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 $R$ such that

$\forall x, y \in I,$ $\forall t,$ $x < t < y$ $⟹$ $t \in I .$

Thus, together with any two points, an interval contains all intermediate points.

It is not difficult to establish the following classification of intervals:

Claim: classification of intervals in $R$ .

$I \subseteq R$ is an interval $⟺$ $I$ is a set of the form $(a, b),$ $[a, b),$ $(a, b],$ $[a, b],$ $(a, + \infty),$ $[a, + \infty),$ $(- \infty, b)$ or $(- \infty, b]$ for some real $a, b,$ or $I = R .$

Remark: the empty subset $\emptyset$ of $R$ 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.

Sketch of proof of the Claim (not given in class). It is clear that every set $(a, b),$ $[a, b), \dots,$ $R,$ listed in the claim, is an interval. Conversely, assume that $I$ is an interval in $R$ 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 $f : X \to R$ 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

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

The set $U \subseteq X$ is open in $X$ because $U$ is the preimage of the set $(- \infty, t),$ open in $R,$ under a continuous function $f .$ Furthermore, $U$ is not empty, as $f (U) ∋ a .$ 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,

$U \cup V = f^{- 1} (R ∖ {t}) = X ∖ f^{- 1} ({t}),$

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 $g : X \to {0, 1}$ be continuous. As in an earlier example, we can view the discrete set ${0, 1}$ as a subspace of the Euclidean line $R .$ The inclusion map $in : {0, 1} \to R$ is continuous by Proposition 2.7, and so we have a continuous map $X \overset{g}{\to} {0, 1} \overset{in}{\to} R .$ By (ii), the image of this map must be an interval in $R;$ 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).

Figure 5.2: defining $g : X \to {0, 1}$ where $X$ is disconnected

(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):

$g : X \to {0, 1}, g (x) = {\begin{cases} 0, & x \in U, \\ 1, & x \in V . \end{cases}$

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 $I \subseteq R$ is connected.

Solution: we use condition (ii) from the Proposition. Let $f : I \to R$ 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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