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FALSE: the projections of a product are closed maps

Statement

False claim: every projection $\pi_j : \prod_i X_i \to X_j$ of a product with the product topology is a closed map, that is, carries closed subsets of the product to closed subsets of $X_j$ (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, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

What is true is the corresponding statement for open sets: every projection is an open map (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 3). The claim above fails already for the binary product $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ , whose product topology is the usual one (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 witness is the hyperbola

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

which is closed in $\mathbb{R}^2$ while $\pi_0[H] = \mathbb{R} \setminus \{0\}$ is not closed in $\mathbb{R}$ .

Facts & Assumptions

Given: $\mathbb{R}^2 = \prod_{k<2}\mathbb{R}$ with the product topology, the first projection $\pi_0(x,y) = x$ , and the set $H$ of the statement.

[A1]

The product topology on $\mathbb{R}^2$ is the metric topology of $d_\infty((x,y),(x',y')) = \max\{|x-x'|,\ |y-y'|\}$ , so $\mathbb{R}^2$ is metrizable (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).

[A2]

$f$ is a closed map when images of closed sets are closed (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[L1]

The multiplication map $m : \mathbb{R}^2 \to \mathbb{R}$ , $m(x,y) := xy$ , is continuous. Indeed, at $(a,b)$ and for $\varepsilon > 0$ , put $\delta := \min\left\{1,\ \frac{\varepsilon}{|a|+|b|+1}\right\}>0.$ If $d_\infty((x,y),(a,b))<\delta$ , then $|x|<|a|+1$ and $|xy-ab|\le |x|\,|y-b|+|b|\,|x-a| < (|a|+|b|+1)\delta\le\varepsilon.$ The bound uses $xy-ab = x(y-b)+b(x-a)$ , the triangle inequality $|u+v|\le|u|+|v|$ (The triangle inequality) and $|uv|=|u|\,|v|$ (Basic properties of the absolute value). This is the metric definition of continuity (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Inverses of positives are positive, and reciprocation reverses order, Maximum and minimum of a set).

[L2]

The singleton $\{1\}$ is closed in $\mathbb{R}$ : if $t\ne1$ , the open interval of radius $|t-1|/2$ about $t$ avoids $1$ . A continuous map of metric spaces has closed preimages of closed sets (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, For a map of metric spaces the following agree: $\varepsilon$ - $\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\overline{A}) \subseteq \overline{f(A)}$ , clause (c)).

[L3]

$U \subseteq \mathbb{R}$ is open in the usual topology exactly when every point of $U$ has a bounded open interval around it inside $U$ ; $(a,b) = \{t : a < t < b\}$ (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).

Refutation

technique · direct

1.1

Since $H = m^{-1}[\{1\}]$ , [L1] and [L2] show that $H$ is closed in $\mathbb{R}^2$ .

A1L1L2

1.2

For $x \ne 0$ the point $(x, 1/x)$ lies in $H$ , since $x \cdot (1/x) = 1$ ; and $(0,y) \notin H$ for every $y$ , since $0 \cdot y = 0 \ne 1$ . So $\pi_0[H] = \mathbb{R} \setminus \{0\}$ .

given

1.3

$\mathbb{R} \setminus \{0\}$ is not closed in $\mathbb{R}$ : its complement $\{0\}$ is not open, because for every $r > 0$ the interval $(-r,r)$ contains $r/2$ , which is nonzero and hence outside $\{0\}$ .

2.1

By step 1.1 the set $H$ is closed in $\mathbb{R}^2$ , while by steps 1.2 and 1.3 its image $\pi_0[H] = \mathbb{R} \setminus \{0\}$ is not closed in $\mathbb{R}$ ; so $\pi_0$ is not a closed map by [A2] and the claim is false.

step 1.1step 1.2step 1.3A2∎

Remarks

The failure is about unboundedness, not about the hyperbola. As $x$ runs to $0$ through positive values the second coordinate $1/x$ runs away, so the points of $H$ approach the vertical axis without meeting it; the image "loses" the point $0$ that the closed set never had. A hypothesis that forbids the escape, compactness of the other factor, is what makes a projection closed, and it is not available at this point in the reading order.
Open and closed are independent for projections. Every projection is open (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), and this item shows that no projection theorem for closed sets follows from it. That is consistent with Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, where an open map that is not closed and a closed map that is not open are exhibited on a two-point space.
The same witness is worked in full on the companion page as The hyperbola $\{(x,y) : xy = 1\}$ is closed in $\mathbb{R}^2$ and its image under the first projection is $\mathbb{R} \setminus \{0\}$ , which is not closed ↗, where the image is computed again and the same choice-free closed-preimage argument establishes that the hyperbola is closed.

Depends on

Used by

The hyperbola {(x,y) : xy = 1} is closed in ℝ² and its image under the first projection is ℝ ∖ {0}, which is not closed Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 106 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Product topology (Wikipedia) (standard reference, not scraped)
Open and closed maps (Wikipedia) (standard reference, not scraped)
Hyperbola (Wikipedia) (standard reference, not scraped)
A quotient map which is neither open nor closed (UC Riverside Math 205A notes) (standard reference, not scraped)