Subspaces, Products, and Quotients: Examples and Counterexamples

Fix $n \in \mathbb{N}$ with $n \ge 1$ and give $\mathbb{R}$ its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Let $\mathbb{R}^n = \prod_{k<n} \mathbb{R}$ carry the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), with projections $\pi_j(x) = x_j$. Then:

1. The product topology is the Euclidean topology. It is the metric topology of $d_\infty$, and equally of $d_1$ and of $d_2$ ($\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space); in particular $\mathbb{R}^n$ with the product topology is metrizable, and "open in $\mathbb{R}^n$" has one meaning.
2. A basis of open boxes. The sets $\prod_{k<n}(a_k, b_k)$ with $a_k < b_k$ for every $k < n$ form a basis (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length), since the $d_\infty$-ball $B(x,r)$ is exactly the box $\prod_{k<n}(x_k - r,\ x_k + r)$.
3. The projections are continuous, open and surjective. Continuity and openness are the general facts (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claims 1 and 3; Continuity of a map of topological spaces at a point and globally, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological). Surjectivity needs no choice principle here: for $t \in \mathbb{R}$ the constant function $x$ with $x_k = t$ for every $k<n$ satisfies $\pi_j(x) = t$.
4. Componentwise continuity. For a space $Z$, a function $h : Z \to \mathbb{R}^n$ is continuous if and only if each of its $n$ components $h_k = \pi_k \circ h : Z \to \mathbb{R}$ is (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 2). This is the statement usually quoted as "a vector-valued map is continuous exactly when its coordinate functions are", and here it is a special case of a theorem about arbitrary products.

• This is the example the seam lemma exists for. Without For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space the symbol $\mathbb{R}^2$ would name two spaces on these two pages, the product of two lines and the metric space of $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$, and $d_1$, $d_2$, $d_\infty$ are metrics on it, and every sentence about open subsets of it would be ambiguous. Claim 1 says the two are one space, so the hyperbola, the square and the Sorgenfrey plane below may each be discussed in whichever language is shorter.

• Surjectivity is free here and is not free in general. For an infinite index set, surjectivity of a projection is the Axiom of Choice (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 4). For $\mathbb{R}^n$ the constant function does the work, and the same trick works for any product of copies of one nonempty space, over any index set.

• The Euclidean metric plays no role in the topology. All three of $d_1$, $d_2$ and $d_\infty$ induce the product topology, so nothing topological about $\mathbb{R}^n$ singles out $d_2$; what singles it out is metric structure, such as which sets are balls, and that is not visible to the topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Refuted: that the projections of a product with the product topology are closed maps (FALSE: the projections of a product are closed maps, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Witness. In $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ with the product topology, which is the usual topology (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space, $\mathbb{R}^n$ as the product of $n$ copies of the real line: the product topology is the Euclidean topology and the projections are continuous, open and surjective), take

$H \;:=\; \{\, (x,y) \in \mathbb{R}^2 : xy = 1 \,\} .$

Then $H$ is closed in $\mathbb{R}^2$, its image under the first projection is $\pi_0[H] = \mathbb{R}\setminus\{0\}$, and $\mathbb{R}\setminus\{0\}$ is not closed in $\mathbb{R}$: the point $0$ lies in its closure and not in it. So $\pi_0$ is not a closed map, although it is a continuous open surjection (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

• The picture behind the computation. The two branches of $H$ run away to infinity as $x$ approaches $0$, so the first coordinates of points of $H$ come arbitrarily close to $0$ while $H$ itself stays away from the whole vertical axis: the point of $H$ nearest to $(0,y_0)$ with $y_0 \ne 0$ is at horizontal distance about $1/|y_0|$, and $H$ approaches the axis only at unbounded heights. Projecting forgets the height, so the first coordinates alone fill $\mathbb{R} \setminus \{0\}$ and their limit point $0$ is missing from the image; nothing about $H$ being closed prevents that, because the points of $H$ whose images converge to $0$ escape to infinity instead of converging in $\mathbb{R}^2$.

• Which hypothesis would repair it, and why it is not available. If the second factor were compact, the projection along it would be a closed map, and $H$ would then have to meet the axis. Compactness is later in the reading order, so no repair is stated here; what is recorded is only the failure.

• The other two conclusions about projections survive untouched. $\pi_0$ is open and surjective (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, $\mathbb{R}^n$ as the product of $n$ copies of the real line: the product topology is the Euclidean topology and the projections are continuous, open and surjective); those hold for every product and are not affected by this witness. Openness and closedness are independent properties of a map, as Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological records.

Refuted: that the box topology has the characteristic property of a product, that is, that a map into $\prod_i X_i$ with all components continuous is continuous for the box topology. Equivalently, this exhibits again that the two topologies differ (FALSE: the product topology and the box topology agree on every product).

Witness. Let $P := \mathbb{R}^{\mathbb{N}} = \prod_{k \in \mathbb{N}}\mathbb{R}$ with every factor carrying the usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and let

$\Delta : \mathbb{R} \to P, \qquad \Delta(t)_k := t \quad (k \in \mathbb{N})$

be the diagonal map. Every component $\pi_k \circ \Delta$ is the identity of $\mathbb{R}$, hence continuous. Then $\Delta$ is continuous for the product topology (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 2) and is not continuous for the box topology: the box

$B \;:=\; \prod_{k \in \mathbb{N}} \Big(-\tfrac{1}{k+1},\ \tfrac{1}{k+1}\Big)$

is box-open and $\Delta^{-1}[B] = \{0\}$, which is not open in $\mathbb{R}$.

• This is the practical reason the product topology is the default. The characteristic property of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice is what makes a map into a product easy to build, and the box topology has no such property: here every component is the identity, and continuity still fails. Any construction that assembles a map coordinate by coordinate would break in the box topology.

• The two topologies are separated by this single map. If they agreed, the same map could not be continuous for one and not for the other; so this item reproves the strictness recorded in FALSE: the product topology and the box topology agree on every product, by a different route and with the same box.

• Nothing here needs a choice principle, the box $B$ and the map $\Delta$ being written down by formulas, and the only existential step being the Archimedean one of For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$.

Let $D := \{0,1\}$ carry the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and let

$K \;:=\; D^{\mathbb{N}} \;=\; \prod_{j \in \mathbb{N}} D$

carry the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Let $C$ be the Cantor set (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) with the subspace topology inherited from the usual topology of $\mathbb{R}$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded). Define

$\Phi : K \to C, \qquad \Phi(b) \;:=\; \sum_{j=0}^{\infty} 2 b_j\, 3^{-(j+1)} .$

Then:

1. $\Phi$ is a well defined bijection onto $C$. Writing $\Phi_0(a) := \sum_{k=0}^{\infty} a_k 3^{-k-1}$ for a sequence $a : \mathbb{N} \to \{0,2\}$, one has $\Phi(b) = \Phi_0\big((2b_k)_k\big)$, and claim 3 of The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$ says exactly that this assignment is a bijection from $\{0,1\}^{\mathbb{N}}$ onto $C$.
2. Two estimates control $\Phi$ completely. For $b, c \in K$ and $n \in \mathbb{N}$:
   • if $b_j = c_j$ for every $j < n$, then $|\Phi(b) - \Phi(c)| \le 3^{-n}$;
   • if $b_j = c_j$ for every $j < m$ and $b_m \ne c_m$, then $|\Phi(b) - \Phi(c)| \ge 3^{-(m+1)}$.
3. $\Phi$ is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological): it is continuous by the first estimate and open onto $C$ by the second, and a continuous open bijection is a homeomorphism (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces).

So the Cantor set, a subspace of the line, and the space of all binary sequences, a product of two-point discrete spaces, are the same topological space; the ternary digits of a point of $C$ are its coordinates in the product.

Compactness is not used anywhere below. The usual argument, that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, is not available at this point in the reading order, and the openness of $\Phi$ is proved by hand instead.

• The two estimates say that $\Phi$ almost preserves distance. Agreement of the first $n$ coordinates forces the images to be within $3^{-n}$, and the first disagreement at index $m$ forces them to be at least $3^{-(m+1)}$ apart. Together they say that the cylinder $Z(b,n)$ and the trace on $C$ of an interval of length about $3^{-n}$ around $\Phi(b)$ determine each other, which is exactly what makes $\Phi$ a homeomorphism.

• Why openness has to be proved and not quoted. For a continuous bijection, openness is equivalent to being a homeomorphism (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces) and is not automatic; the standard shortcut uses compactness of the source and the Hausdorff condition on the target, and compactness is later in the reading order. Steps 4.3 and 5.1 replace it with a direct computation.

• The coordinates are the digits, and the digits are not the point. A real number in $C$ has exactly one ternary expansion with digits in $\{0,2\}$, which is what makes $\Phi$ injective; the ambiguity of ternary expansions in general, such as two expansions of $1/3$, does not arise inside $C$ because the alternative expansion uses the digit $1$.

Let $[0,1]$ carry the subspace topology from the usual topology of $\mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded) and let

$Q \;:=\; [0,1]^{\mathbb{N}} \;=\; \prod_{k \in \mathbb{N}} [0,1]$

carry the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space); $Q$ is the Hilbert cube. Define

$d(x,y) \;:=\; \sum_{k=0}^{\infty} \frac{|x_k - y_k|}{2^{\,k+1}} \qquad (x,y \in Q).$

Then:

1. $d$ is defined, with $0 \le d(x,y) \le 1$: the series has nonnegative terms bounded by $2^{-(k+1)}$, and $\sum_{k} 2^{-(k+1)} = 1$ (For $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ the series diverges, If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$, and divergence of $\sum a_k$ gives divergence of $\sum b_k$).
2. $d$ is a metric on $Q$ (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric).
3. $d$ induces the product topology, so $Q$ is metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Claim 3 is the only real work, and it is what the index weights are for: the factor $2^{-(k+1)}$ makes the tail of the sum small no matter what the coordinates do, so a constraint on finitely many coordinates already forces $d$ to be small, and conversely $d$ small forces each individual coordinate to be close.

By claim 1 of Products commute with subspaces; for infinite nonempty families, the closure identity $\overline{\prod A_i}=\prod \overline{A_i}$ uses the Axiom of Choice the topology on $Q$ is also the subspace topology it inherits from $\mathbb{R}^{\mathbb{N}}$, so the two readings of "$[0,1]^{\mathbb{N}}$" agree.