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 $f\colon Z \to X \times Y$ 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

\[ f_X : Z \xrightarrow {f} X\times Y \xrightarrow {p_X} X, \quad f_Y : Z \xrightarrow {f} X\times Y \xrightarrow {p_Y} Y \]

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 $f\colon Z\to X \times Y,$ and
• pairs of continuous functions $(f_X\colon Z\to X, f_Y\colon Z\to Y).$

To $f\colon Z\to X \times Y$ 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 $f\colon Z \to X \times Y$ 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 $f\colon Z \to X \times Y$ 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 $f_X\colon Z \to X,$ $f_Y\colon Z \to Y$ 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 $\bigcup _{\alpha \in I}U_\alpha \times V_\alpha $ of open rectangles, where, for each $\alpha \in I,$ $U_\alpha $ is open in $X$ and $V_\alpha $ is open in $Y.$ The preimage of a union is the union of preimages, so

\[ f^{-1}\bigl (\bigcup _{\alpha \in I}U_\alpha \times V_\alpha \bigr ) = \bigcup _{\alpha \in I} f^{-1}(U_\alpha \times V_\alpha ). \]

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

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

It remains to note that, given $f\colon Z \to X \times Y,$ 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 $i_{y_0}\colon X \to X \times Y,$ $x\to (x,y_0).$ 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 $\mathrm {const}_{y_0}\colon X \to Y,$ 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 $p_X \colon X \times Y \to X$ 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 $i_{y_0}\colon X \to X\times \{y_0\}.$ □

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

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

If $x\ne x',$ 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 $y\ne y',$ 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 $\mathscr B,$ we first prove a lemma which allows us to check compactness using only covers by sets from $\mathscr B.$

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

Let $\mathscr B$ be a base of topology on a space $X.$ Suppose that every basic cover of $X$ (i.e., a cover by sets from $\mathscr B$) 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 $\mathscr B$). Call these basic open sets “the children” of $U.$

Suppose that $\mathscr C$ is a cover of $X$ by open sets $U_\alpha ,$ $\alpha \in I.$ Consider the collection $\mathscr C_1 = \{$all children of all sets from $\mathscr C\}.$ Each set in $\mathscr C$ is the union of its children, so $\bigcup \mathscr C_1 = \bigcup \mathscr C = X.$

Thus, $\mathscr C_1$ is a basic open cover of $X.$ Then by assumption, a finite cover $V_1,\dots , V_n$ can be chosen from $\mathscr C_1:$ $V_1\cup \dots \cup V_n= X.$ The sets $V_1,\dots ,V_n$ may not lie in $\mathscr C,$ but their “parents” do. Consider the finite subcollection of $\mathscr C$ given by

a parent of $V_1,$ a parent of $V_2,$ $\ldots ,$ 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 $\mathscr C$ 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 $\iff $ $X\times Y$ 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 $\mathscr C$ is a cover of $X\times Y$ by open rectangles and show that $\mathscr C$ 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 $\mathscr C.$ We may assume that each $U_i\times V_i$ intersects $X\times \{y_0\}$ (otherwise delete it from the finite cover), so $V_i\ni y_0$ 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 $\mathscr C.$ The union of these $m$ finite subcollections is a finite subcover, chosen from $\mathscr C,$ 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 $\RR ^n.$

Theorem 9.6: The Heine-Borel Theorem.

In the Euclidean space $\RR ^n,$ 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 $\RR ^n.$ Any bounded set is a subset of an $n$-dimensional cube $\{(x_1,\dots ,x_n)\in \RR ^n: |x_i|\le M\ \forall i=1,\dots ,n\},$ for some $M>0.$ This cube of side length $2M$ is the product space $[-M,M]\times \dots \times [-M,M] = [-M,M]^n.$

Note that the closed bounded interval $[-M,M]\subseteq \RR $ 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 $\RR ^3$

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 $f\colon X \to Y$ such that restricting the codomain gives a homeomorphism $f\colon X \xrightarrow {\sim } f(X).$

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 $\RR ^3$.

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

\[ \mathbb T^2 = S^1 \times S^1. \]

Show: $\mathbb T^2$ is compact. Construct an embedding of $\mathbb T^2$ in the Euclidean space $\RR ^3.$

Solution: $S^1$ is closed and bounded in $\RR ^2,$ 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 $[0,1]\to \RR ^2,$ $t\mapsto (\cos 2\pi t, \sin 2\pi t)$ which is continuous; $[0,1]$ is compact, and a continuous image of a compact is compact.)

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

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

A point of $\mathbb T^2$ 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 $(\varphi , \theta )$ with $\varphi ,\theta \in [0,2\pi ),$ 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 $xy$ plane in $\RR ^3.$ The torus $\mathbb T^2$ is the disjoint union

\[ \mathbb T^2 = \bigcup _{P\in S^1} \{P\}\times S^1, \]

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, $f\colon S^1\times S^1 \to \RR ^3$ is given by

\[ (\varphi , \theta ) \mapsto \textrm {Rotate}^\varphi _{\text {$z$-axis}} \bigl ( (R,0,0) + (\cos \theta , 0, \sin \theta )\bigr ) \]

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

Explanation why $f$ is injective: $f$ maps two points $(\varphi ,\theta )$ and $(\varphi ',\theta ')$ on $\mathbb T^2$ such that $\varphi \ne \varphi ',$ 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 $\varphi =\varphi ',$ the two points are mapped by $f$ onto the same red circle but to different points of the circle, as long as $\theta \ne \theta '.$

Explanation why $f$ is continuous: the $x$-component $f_x$ of $f$ can be written as an algebraic expression in $\cos \varphi ,$ $\sin \varphi ,$ $\cos \theta $ and $\sin \theta $ (see the formula given in the tutorial). Note that $\cos \varphi $ and $\sin \varphi $ are the actual coordinates of the point on the first (blue) circle, hence they are continuous functions on $\mathbb T^2$ by Proposition 8.1. Same can be said of $\cos \theta $ and $\sin \theta .$ 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 $\mathbb T^2$ to $\RR .$

In the same way one shows that $f_y$ and $f_z$ are continuous $\RR $-valued functions on $\mathbb T^2.$ Hence $f\colon \mathbb T^2 \to \RR ^3$ 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 $f\colon \mathbb T^2 \to f(\mathbb T^2).$ We do not need continuity of the inverse map: as $\mathbb T^2$ is compact and $f(\mathbb T^2)$ is metric hence Hausdorff, by the Topological Inverse Function Theorem 4.5 $f\colon \mathbb T^2 \xrightarrow {\sim } f(\mathbb T^2)$ 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 $(X_\alpha )_{\alpha \in I}$ be a collection of topological spaces. The topology on the Cartesian product $\prod _{\alpha \in I} X_\alpha $ is defined to have base $\mathscr B$ of sets of the form $\prod _{\alpha \in I} U_\alpha $ where (1) $U_\alpha = X_\alpha $ for all but finitely many $\alpha \in I;$ (2) $U_\alpha $ is open in $X_\alpha $ for all $\alpha \in I.$

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

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 $i_{y_0}\colon x\mapsto (x,y_0)$ 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].

Version 2024/12/03 Week 9 in PDF All notes in PDF To other weeks

Week 9 Topological properties of product spaces

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

Topological properties of the product space

The baby Tychonoff theorem about compactness of \(X\times Y\)

The Heine-Borel Theorem

The torus and its embedding in \(\RR ^3\)

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

References for the week 9 notes