Week 9 notes

Week 9 Topological properties of product spaces

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The universal mapping property of the product space $X \times Y$

Any function $f : Z \to X \times Y$ must output a point of $X \times Y$ which is a pair, hence must be of the form $f (z) = (f_{X} (z), f_{Y} (z)) .$ Formally, the “components” of $f$ are

$f_{X} : Z \overset{f}{\to} X \times Y \overset{p_{X}}{\to} X, f_{Y} : Z \overset{f}{\to} X \times Y \overset{p_{Y}}{\to} Y$

where $p_{X}, p_{Y}$ are the projections. The next result is called the universal mapping property because it describes all possible continuous functions with the codomain $X \times Y .$

Theorem 9.1: the universal mapping property.

Let $X, Y, Z$ be topological spaces. There is a 1-to-1 correspondence between

• continuous functions $f : Z \to X \times Y,$ and
• pairs of continuous functions $(f_{X} : Z \to X, f_{Y} : Z \to Y) .$

To $f : Z \to X \times Y$ there corresponds the pair $(f_{X} = p_{X} \circ f, f_{Y} = p_{Y} \circ f) .$ Vice versa, to a pair $(f_{X}, f_{Y})$ there corresponds the function $f$ defined by $f (z) = (f_{X} (z), f_{Y} (z)) .$

Remark: put simply, the Theorem says that $f : Z \to X \times Y$ is continuous if and only if the functions $f_{X} = p_{X} \circ f,$ $f_{Y} = p_{Y} \circ f$ are continuous.

Proof of the Theorem. Let $f : Z \to X \times Y$ be a continuous function. Since $p_{X}, p_{Y}$ are continuous by Proposition 8.1, and a composition of continuous functions is continuous, the functions $f_{X} = p_{X} \circ f,$ $f_{Y} = p_{Y} \circ f$ are continuous.

Vice versa, let pair $f_{X} : Z \to X,$ $f_{Y} : Z \to Y$ be continuous functions. We need to check that the function $f = (f_{X}, f_{Y})$ is continuous. By definition of the product topology, an arbitrary open subset of $X \times Y$ is a union $⋃_{𝛼 \in I} U_{𝛼} \times V_{𝛼}$ of open rectangles, where, for each $𝛼 \in I,$ $U_{𝛼}$ is open in $X$ and $V_{𝛼}$ is open in $Y .$ The preimage of a union is the union of preimages, so

$f^{- 1} (⋃_{𝛼 \in I} U_{𝛼} \times V_{𝛼}) = ⋃_{𝛼 \in I} f^{- 1} (U_{𝛼} \times V_{𝛼}) .$

Note that $f^{- 1} (U_{𝛼} \times V_{𝛼})$ consists of $z \in Z$ such that $f_{X} (z) \in U_{𝛼}$ and $f_{Y} (z) \in V_{𝛼} .$ In other words, $f^{- 1} (U_{𝛼} \times V_{𝛼}) = f_{X}^{- 1} (U_{𝛼}) \cap f_{Y}^{- 1} (V_{𝛼})$ which is open in $Z,$ because $f_{X}^{- 1} (U_{𝛼})$ and $f_{Y}^{- 1} (V_{𝛼})$ are preimages of open sets under the confinuous functions $f_{X}, f_{Y},$ and the intersection of two open sets is open.

Therefore, $⋃_{𝛼 \in I} f^{- 1} (U_{𝛼} \times V_{𝛼})$ is open as a union of open sets. We have verified the definition of “continuous” for $f .$

It remains to note that, given $f : Z \to X \times Y,$ taking $f_{X} = p_{X} \circ f$ and $f_{Y} = p_{Y} \circ f$ then constructing $(f_{X}, f_{Y})$ brings us back to the function $f .$ Also, given the functions $f_{X},$ $f_{Y},$ if $f = (f_{X}, f_{Y})$ then taking $p_{X} \circ f$ and $p_{Y} \circ f$ returns us to the functions $f_{X}$ and $f_{Y} .$ This shows that we have two mutually inverse correspondences between functions $Z \to X \times Y$ and pairs of functions $Z \to X,$ $Z \to Y .$ Hence we have a bijective (1-to-1) correspondence between the two sets of functions, as claimed. □

Topological properties of the product space

We would like to understand how the topological properties of the product space $X \times Y$ are determined by the topological properties of the spaces $X$ and $Y .$ The following result helps us by showing how homeomorphic copies of the space $X$ sit inside $X \times Y :$

Lemma 9.2: embedding of $X$ in $X \times Y$ .

For each $y_{0} \in Y,$ the subspace $X \times {y_{0}}$ of $X \times Y$ is homeomorphic to $X .$

Proof. Consider the embedding map $i_{y_{0}} : X \to X \times Y,$ $x \to (x, y_{0}) .$ Note that $p_{X} \circ i_{y_{0}}$ is the identity map $x \mapsto x$ of $X$ which is continuous, and $p_{Y} \circ i_{y_{0}}$ is the constant map ${const}_{y_{0}} : X \to Y,$ which is continuous. Hence by the Universal Mapping Property, Theorem 9.1, $i_{y_{0}}$ is a continuous map. We can restrict the codomain and consider $i_{y_{0}}$ as a continuous map $X \to X \times {y_{0}} .$

The projection $p_{X} : X \times Y \to X$ is continuous by Proposition 8.1, so its restriction $p_{X} |_{X \times {y_{0}}}$ is continuous.

The composition $x \mapsto (x, y_{0}) \mapsto x$ shows that $p_{X} |_{X \times {y_{0}}} \circ i_{y_{0}} = {id}_{X} .$ Also, the composition $(x, y_{0}) \mapsto x \mapsto (x, y_{0})$ shows that $i_{y_{0}} \circ p_{X} |_{X \times {y_{0}}} = {id}_{X \times {y_{0}}} .$ Therefore, $i_{y_{0}}$ and $p_{X} |_{X \times {y_{0}}}$ are two mutually inverse continuous maps. We have verified the definition of “homeomorphism” for $i_{y_{0}} : X \to X \times {y_{0}} .$ □

Remark: in the same way, if $x_{0} \in X,$ the subspace ${x_{0}} \times Y$ of $X \times Y$ is homeomorphic to $Y .$

Proposition 9.3: Hausdorfness and connectedness of the product.

If $X, Y$ are non-empty topological spaces, then

(a) $X$ and $Y$ are both Hausdorff $⟺$ $X \times Y$ is Hausdorff;
(b) $X$ and $Y$ are both connected $⟺$ $X \times Y$ is connected.

Proof (not given in class). (a) $\Rightarrow:$ assume $X,$ $Y$ are Hausdorff, and let $(x, y) \neq (x^{'}, y^{'})$ be two distinct points of $X \times Y .$ Then either $x \neq x^{'},$ or $y \neq y^{'},$ or both.

If $x \neq x^{'},$ take $U, U^{'}$ to be open sets in $X$ such that $x \in U,$ $x^{'} \in U^{'}$ and $U \cap U^{'} = \emptyset .$ Then the points $(x, y)$ and $(x^{'}, y^{'})$ lie in disjoint open cylinder sets $U \times Y$ and $U^{'} \times Y .$ If $y \neq y^{'},$ a similar argument leads to disjoint open cylinders $X \times V$ and $X \times V^{'} .$ We have thus verified the definition of “Hausdorff” for $X \times Y .$

$\Leftarrow:$ assume $X \times Y$ is Hausdorff. Pick a point $y_{0}$ in the non-empty space $Y .$ Subspaces of a Hausdorff space are Hausdorff (Proposition 3.3), so $X \times {y_{0}}$ is Hausdorff; it is homeomorphic to $X$ by Lemma 9.2, and Hausdorffness is a topological property (Proposition 3.1), so $X$ is Hausdorff. Similarly, $Y$ is Hausdorff.

Figure 9.1: if $X,$ $Y$ are connected, any two points of $X \times Y$ are joined by a connected set

(b) $\Rightarrow:$ assume $X,$ $Y$ are connected. Any two points $(x, y)$ and $(x^{'}, y^{'})$ of $X \times Y$ lie in the set $(X \times {y}) \cup ({x^{'}} \times Y),$ see Figure 9.1. Connectedness is a topological property, so $X \times {y},$ which is homeomorphic to $X$ by Lemma 9.2, is connected. Similarly, ${x^{'}} \times Y$ is connected. The union of two connected sets, which have a common point, is connected by Lemma 7.3, so $(x, y)$ and $(x^{'}, y^{'})$ lie in the same connected component of $X \times Y .$ Since the two points were arbitrary, $X \times Y$ has only one connected component, i.e., is connected.

$\Leftarrow:$ if $X \times Y$ is connected, then $X = p_{X} (X \times Y)$ and $Y = p_{Y} (X \times Y)$ are connected, because $p_{X},$ $p_{Y}$ are continuous (Proposition 8.1) and a continuous image of a connected space is connected (Theorem 7.1). □

The baby Tychonoff theorem about compactness of $X \times Y$

We will now deal with compactness of $X \times Y$ which is more intricate than Hausdorfness and connectedness. Since the product topology on $X \times Y$ is given by its base $B,$ we first prove a lemma which allows us to check compactness using only covers by sets from $B .$

Lemma 9.4: a “basic compact” is a compact.

Let $B$ be a base of topology on a space $X .$ Suppose that every basic cover of $X$ (i.e., a cover by sets from $B$ ) has a finite subcover. Then $X$ is compact.

Proof. By definition of a base, every open set $U \subseteq X$ can be written as a union of basic open sets (i.e., members of $B$ ). Call these basic open sets “the children” of $U .$

Suppose that $C$ is a cover of $X$ by open sets $U_{𝛼},$ $𝛼 \in I .$ Consider the collection $C_{1} = {$ all children of all sets from $C} .$ Each set in $C$ is the union of its children, so $⋃ C_{1} = ⋃ C = X .$

Thus, $C_{1}$ is a basic open cover of $X .$ Then by assumption, a finite cover $V_{1}, \dots, V_{n}$ can be chosen from $C_{1} :$ $V_{1} \cup \dots \cup V_{n} = X .$ The sets $V_{1}, \dots, V_{n}$ may not lie in $C,$ but their “parents” do. Consider the finite subcollection of $C$ given by

a parent of $V_{1},$ a parent of $V_{2},$ $\dots,$ a parent of $V_{n} .$

Here “a parent” means an arbitrary choice of parent if $V_{i}$ has more than one parent. Since $V_{i} \subseteq ($ a parent of $V_{i}),$ the union of the $n$ “parents” is also $X .$ Given any open cover $C$ of $X,$ we constructed a finite subcover, thus verifying the definition of “compact” for $X .$ □

Theorem 9.5: the baby Tychonoff theorem.

If $X, Y$ are non-empty spaces, both $X$ and $Y$ are compact $⟺$ $X \times Y$ is compact.

(rectangles covering a horizontal line inside the square X times Y)

Proof. $\Leftarrow:$ $X \times Y$ is compact, so $X = p_{X} (X \times Y)$ is compact by Theorem 4.2 (continuous image of compact) as $p_{X}$ is continuous by Proposition 8.1. Similarly, $Y$ is compact.

$\Rightarrow:$ let $X,$ $Y$ be compact. We assume that $C$ is a cover of $X \times Y$ by open rectangles and show that $C$ has a finite subcover. Since, by definition, open rectangles form a base of the topology on $X \times Y,$ Lemma 9.4 will imply that $X \times Y$ is compact.

For each $y_{0} \in Y,$ $X \times {y_{0}}$ is homeomorphic to $X$ (Lemma 9.2), hence $X \times {y_{0}}$ is a compact set, which by compactness criterion 4.1 is covered by a finite subcollection of open rectangles $U_{1} \times V_{1}, \dots, U_{n} \times V_{n}$ from $C .$ We may assume that each $U_{i} \times V_{i}$ intersects $X \times {y_{0}}$ (otherwise delete it from the finite cover), so $V_{i} ∋ y_{0}$ for all $i = 1, \dots, n .$

Define the open neighbourhood $V (y_{0})$ to be $V_{1} \cap \dots \cap V_{n} .$ Then each open rectangle $U_{i} \times V_{i}$ contains $U_{i} \times V (y_{0}),$ and so the open rectangles $U_{1} \times V_{1}, \dots, U_{n} \times V_{n},$ which form the finite cover of $X \times {y_{0}},$ also cover the open cylinder $X \times V (y_{0}),$ see Figure 9.2.

We have thus constructed an open neighbourhood $V (y_{0})$ for every point $y_{0}$ of $Y .$ We now use compactness of $Y$ to choose a finite subcover of $Y$ by these neighbourhoods: say, $V (y_{1}), \dots, V (y_{m})$ such that $V (y_{1}) \cup \dots \cup V (y_{m}) = Y .$

Then the union of the cylinders $X \times V (y_{1}), \dots, X \times V (y_{m})$ is $X \times Y .$ Also, by the above construction, each cylinder $X \times V (y_{j})$ is covered by a finite subcollection of $C .$ The union of these $m$ finite subcollections is a finite subcover, chosen from $C,$ for the whole of $X \times Y,$ as required. □

The following corollary is now easily obtained by induction. We understand the $n$ -fold product $X_{1} \times X_{2} \times \dots \times X_{n}$ as the iterated product $(((X_{1} \times X_{2}) \times X_{3}) \times \dots) \times X_{n} .$

Corollary: the product of finitely many compact spaces is compact.

If $X_{1}, \dots, X_{n}$ are compact topological spaces, then the product space $X_{1} \times \dots \times X_{n}$ is compact.□

The Heine-Borel Theorem

The baby Tychonoff theorem is now used to extend the Heine-Borel Lemma, Theorem 5.2, which says that $[0, 1]$ is compact, to a result which describes all compact sets in Euclidean spaces $R^{n} .$

Theorem 9.6: The Heine-Borel Theorem.

In the Euclidean space $R^{n},$ a set $K$ is compact iff $K$ is closed and bounded.

Proof. The “only if” part is immediate by Proposition 5.1: if $K$ is a compact set in any metric space, then $K$ is closed and bounded.

To prove the “if” part, assume that $K$ is a closed bounded subset of $R^{n} .$ Any bounded set is a subset of an $n$ -dimensional cube ${(x_{1}, \dots, x_{n}) \in R^{n} : | x_{i} | \leq M \forall i = 1, \dots, n},$ for some $M > 0.$ This cube of side length $2 M$ is the product space $[- M, M] \times \dots \times [- M, M] = [- M, M]^{n} .$

Note that the closed bounded interval $[- M, M] \subseteq R$ is homeomorphic to $[0, 1]$ (a homeomorphism is afforded, for example, by a linear function mapping $[0, 1]$ onto $[- M, M]$ ). By the Heine-Borel Lemma, $[0, 1]$ is compact, and so $[- M, M]$ is also compact. By baby Tychonoff theorem 9.5 and its Corollary, $[- M, M]^{n}$ is compact.

Thus $K$ is a closed subset of the compact $[- M, M]^{n},$ and so by Proposition 4.3, $K$ is compact. □

The torus and its embedding in $R^{3}$

The theory that we have developed so far can be used to construct embeddings of abstractly defined topological spaces in a Euclidean space. By an embedding, we mean the following:

Definition: embedding.

Let $X,$ $Y$ be topological spaces. An embedding of $X$ in $Y$ is a map $f : X \to Y$ such that restricting the codomain gives a homeomorphism $f : X \tilde{\to} f (X) .$

In other words, an embedding means constructing inside $Y$ a subspace homeomorphic to $X .$ This is important in many applications.

Example: embedding of the torus in $R^{3}$ .

Let $S^{1}$ denote the unit circle ${(x, y) \in R^{2} : x^{2} + y^{2} = 1}$ in the Euclidean plane. Define the torus as the product space

$T^{2} = S^{1} \times S^{1} .$

Show: $T^{2}$ is compact. Construct an embedding of $T^{2}$ in the Euclidean space $R^{3} .$

Solution: $S^{1}$ is closed and bounded in $R^{2},$ hence compact by Heine-Borel Theorem 9.6.

(There are alternative ways to show compactness of $S^{1};$ for example, $S^{1}$ is the image of the map $[0, 1] \to R^{2},$ $t \mapsto (\cos 2 𝜋 t, \sin 2 𝜋 t)$ which is continuous; $[0, 1]$ is compact, and a continuous image of a compact is compact.)

It follows that $T^{2} = S^{1} \times S^{1}$ is compact by baby Tychonoff theorem 9.5.

It is not automatic that an embedding of $T^{2}$ in $R^{3}$ must exist: note that $S^{1}$ is a subspace of $R^{2},$ so $T^{2}$ is naturally a subspace of $R^{2} \times R^{2} = R^{4}$ and not $R^{3} .$ Yet we can construct an injective function $f : S^{1} \times S^{1} \to R^{3},$ as follows.

A point of $T^{2}$ is a pair $(P, Q)$ where $P$ is a point on the first circle, and $Q$ is a point on the second unit circle in $S^{1} \times S^{1} .$ It is convenient to represent the two points by their angle coordinates $(𝜑, 𝜃)$ with $𝜑, 𝜃 \in [0, 2 𝜋),$ see Figure 9.3.

Fix two radii $R, r$ such that $R > r > 0.$ Informally, we will think of the first circle in $S^{1} \times S^{1}$ (the blue circle in Figure 9.3) as the circle of radius $R$ in the horizontal $x y$ plane in $R^{3} .$ The torus $T^{2}$ is the disjoint union

$T^{2} = ⋃_{P \in S^{1}} {P} \times S^{1},$

and for each $P$ on the blue circle, we map the subset ${P} \times S^{1}$ of the torus to the red circle of radius $R,$ centred at $P$ and orthogonal to the blue circle. Thus, $f : S^{1} \times S^{1} \to R^{3}$ is given by

$(𝜑, 𝜃) \mapsto {Rotate}_{z -axis}^{𝜑} ((R, 0, 0) + (\cos 𝜃, 0, \sin 𝜃))$

as shown in Figure 9.4. An explicit formula for $f (𝜑, 𝜃)$ will be worked out in the tutorial.

Explanation why $f$ is injective: $f$ maps two points $(𝜑, 𝜃)$ and $(𝜑^{'}, 𝜃^{'})$ on $T^{2}$ such that $𝜑 \neq 𝜑^{'},$ onto two disjoint red circles in Figure 9.4, to the $f$ -images are distinct. The red circles are disjoint because we chose $r < R .$ In the case $𝜑 = 𝜑^{'},$ the two points are mapped by $f$ onto the same red circle but to different points of the circle, as long as $𝜃 \neq 𝜃^{'} .$

Explanation why $f$ is continuous: the $x$ -component $f_{x}$ of $f$ can be written as an algebraic expression in $\cos 𝜑,$ $\sin 𝜑,$ $\cos 𝜃$ and $\sin 𝜃$ (see the formula given in the tutorial). Note that $\cos 𝜑$ and $\sin 𝜑$ are the actual coordinates of the point on the first (blue) circle, hence they are continuous functions on $T^{2}$ by Proposition 8.1. Same can be said of $\cos 𝜃$ and $\sin 𝜃 .$ Sums and products of continuous functions are continuous (this is known to be true for metric spaces; for general topological spaces, see E4.5). Hence $f_{x}$ is a continuous function from $T^{2}$ to $R .$

In the same way one shows that $f_{y}$ and $f_{z}$ are continuous $R$ -valued functions on $T^{2} .$ Hence $f : T^{2} \to R^{3}$ is continuous by the Universal Mapping Property, Theorem 9.1.

Proof that $f$ is an embedding: restricting the codomain of continuous injection $f$ gives the the continuous bijection $f : T^{2} \to f (T^{2}) .$ We do not need continuity of the inverse map: as $T^{2}$ is compact and $f (T^{2})$ is metric hence Hausdorff, by the Topological Inverse Function Theorem 4.5 $f : T^{2} \tilde{\to} f (T^{2})$ is a homeomorphism. □

The Tychonoff theorem (not done in class, not examinable)

Baby Tychonoff theorem 9.5 is a particular case of a result that we state here for completeness. The proof is beyond the scope of this course and can be found in the literature.

Let $(X_{𝛼})_{𝛼 \in I}$ be a collection of topological spaces. The topology on the Cartesian product $\prod_{𝛼 \in I} X_{𝛼}$ is defined to have base $B$ of sets of the form $\prod_{𝛼 \in I} U_{𝛼}$ where (1) $U_{𝛼} = X_{𝛼}$ for all but finitely many $𝛼 \in I;$ (2) $U_{𝛼}$ is open in $X_{𝛼}$ for all $𝛼 \in I .$

The product topology on $X \times Y$ is a particular case: (1) can be omitted if $I$ is finite.

Theorem 9.7: the Tychonoff theorem.

Suppose the space $X_{𝛼}$ is not empty for all $𝛼 \in I .$ Then the product space $\prod_{𝛼 \in I} X_{𝛼},$ defined above, is compact if, and only if, $X_{𝛼}$ is compact for all $𝛼 \in I .$ □

References for the week 9 notes

Theorem 9.1, the Universal Mapping Property of $X \times Y,$ is [Sutherland, Proposition 10.11] as well as [Armstrong, Theorem (3.13)].

Lemma 9.2 about a homeomorphic copy $X \times {y_{0}}$ of $X$ in $X \times Y$ is a strengthening of [Sutherland, Proposition 10.14] which only asserts that the map $i_{y_{0}} : x \mapsto (x, y_{0})$ is continuous.

Proposition 9.3: (a) Hausdorffness of $X \times Y$ is [Sutherland, Proposition 11.17b], [Armstrong, Theorem (3.14)]; (b) connectedness is [Sutherland, Theorem 12.18], [Armstrong, Theorem (3.26)].

The baby Tychonoff theorem 9.5 is [Sutherland, Theorem 13.21] and [Armstrong, Theorem (3.15)]. Our proof follows [Armstrong]. In particular, our Lemma 9.4 is [Armstrong, Lemma (3.16)].

The Heine-Borel theorem 9.6 is [Sutherland, Theorem 13.22] and [Armstrong, Theorem (3.1)].

The general Tychonoff theorem 9.7 is not usually proved in introductory-level Topology textbooks. A proof can be found in [Willard, Theorem 17.8].