4.11: Product Spaces. Double and Iterated Limits
This page is a draft and is under active development.
Given two metric spaces \(\left(X, \rho_{1}\right)\) and \(\left(Y, \rho_{2}\right),\) we may consider the Cartesian product \(X \times Y,\) suitably metrized. Two metrics for \(X \times Y\) are suggested in Problem 10 in Chapter 3, §11. We shall adopt the first of them as follows.
By the product of two metric spaces \(\left(X, \rho_{1}\right)\) and \(\left(Y, \rho_{2}\right)\) is meant the space \((X \times Y, \rho),\) where the metric \(\rho\) is defined by
\[
\rho\left((x, y),\left(x^{\prime}, y^{\prime}\right)\right)=\max \left\{\rho_{1}\left(x, x^{\prime}\right), \rho_{2}\left(y, y^{\prime}\right)\right\}
\]
for \(x, x^{\prime} \in X\) and \(y, y^{\prime} \in Y\).
Thus the distance between \((x, y)\) and \(\left(x^{\prime}, y^{\prime}\right)\) is the larger of the two distances
\[
\rho_{1}\left(x, x^{\prime}\right) \text { in } X \text { and } \rho_{2}\left(y, y^{\prime}\right) \text { in } Y.
\]
The verification that \(\rho\) in \((1)\) is, indeed, a metric is left to the reader. We now obtain the following theorem.
(i) A globe \(G_{(p, q)}(\varepsilon)\) in \((X \times Y, \rho)\) is the Cartesian product of the corresponding \(\varepsilon\) -globes in \(X\) and \(Y\),
\[
G_{(p, q)}(\varepsilon)=G_{p}(\varepsilon) \times G_{q}(\varepsilon).
\]
(ii) Convergence of sequences \(\left\{\left(x_{m}, y_{m}\right)\right\}\) in \(X \times Y\) is componentwise. That is, we have
\[
\left(x_{m}, y_{m}\right) \rightarrow(p, q) \text { in } X \times Y \text { iff } x_{m} \rightarrow p \text { in } X \text { and } y_{m} \rightarrow q \text { in } Y.
\]
- Proof
-
Again, the easy proof is left as an exercise.
In this connection, recall that by Theorem 2 of Chapter 3, §15, convergence in \(E^{2}\) is componentwise as well, even though the standard metric in \(E^{2}\) is not the product metric \((1) ;\) it is rather the metric (ii) of Problem 10 in Chapter 3, §11. We might have adopted this second metric for \(X \times Y\) as well. Then part (i) of Theorem 1 would fail, but part (ii) would still follow by making
\[
\rho_{1}\left(x_{m}, p\right)<\frac{\varepsilon}{\sqrt{2}} \text { and } \rho_{2}\left(y_{m}, q\right)<\frac{\varepsilon}{\sqrt{2}}.
\]
It follows that, as far as convergence is concerned, the two choices of \(\rho\) are equivalent.
Note 1. More generally, two metrics for a space \(S\) are said to be equivalent iff exactly the same sequences converge (to the same limits) under both metrics. Then also all function limits are the same since they reduential limits, by Theorem 1 of §2; similarly for such notions as continuity, compactness, completeness, closedness, openness, etc.
In view of this, we shall often call \(X \times Y\) a product space (in the wider sense) even if its metric is not the \(\rho\) of formula \((1)\) but equivalent to it. In this sense, \(E^{2}\) is the product space \(E^{1} \times E^{1},\) and \(X \times Y\) is its generalization.
Various ideas valid in \(E^{2}\) extend quite naturally to \(X \times Y .\) Thus functions defined on a set \(A \subseteq X \times Y\) may be treated as functions of two variables \(x,\) \(y\) such that \((x, y) \in A .\) Given \((p, q) \in X \times Y,\) we may consider ordinary or relative limits at \((p, q),\) e.g. limits over a path
\[
B=\{(x, y) \in X \times Y | y=q\}
\]
(briefly called the "line \(y=q^{\prime \prime} ) .\) In this case, \(y\) remains fixed \((y=q)\) while \(x \rightarrow p ;\) we then speak of limits and continuity in one variable \(x,\) as opposed to those in both variables jointly, i.e., the ordinary limits (cf. §3, part \(\mathrm{IV} )\).
Some other kinds of limits are to be defined below. For simplicity, we consider only functions \(f :(X \times Y) \rightarrow\left(T, \rho^{\prime}\right)\) defined on all of \(X \times Y .\) If confusion is unlikely, we write \(\rho\) for all metrics involved (such as \(\rho^{\prime}\) in \(T ) .\) Below, \(p\) and \(q\) always denote cluster points of \(X\) and \(Y,\) respectively (this justifies the "lim" notation. Of course, our definitions apply in particular to \(E^{2}\) as the simplest special case of \(X \times Y\).
A function \(f :(X \times Y) \rightarrow\left(T, \rho^{\prime}\right)\) is said to have the double limit \(s \in T\) at \((p, q),\) denoted
\[
s=\lim _{x \rightarrow p \atop y \rightarrow q} f(x, y),
\]
iff for each \(\varepsilon>0,\) there is a \(\delta>0\) such that \(f(x, y) \in G_{s}(\varepsilon)\) whenever \(x \in G_{\neg p}(\delta)\) and \(y \in G_{\neg q}(\delta) .\) In symbols,
\[
(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in G_{\neg p}(\delta)\right)\left(\forall y \in G_{-q}(\delta)\right) \quad f(x, y) \in G_{s}(\varepsilon).
\]
Observe that this is the relative limit over the path
\[
D=(X-\{p\}) \times(Y-\{q\})
\]
excluding the two "lines" \(x=p\) and \(y=q\). If \(f\) were restricted to \(D,\) this would coincide with the ordinary nonrelative limit (see §1), denoted
\[
s=\lim _{(x, y) \rightarrow(p, q)} f(x, y),
\]
where only the point \((p, q)\) is excluded. Then we would have
\[
(\forall \varepsilon>0)(\exists \delta>0)\left(\forall(x, y) \in G_{\neg(p, q)}(\delta)\right) \quad f(x, y) \in G_{s}(\varepsilon).
\]
Now consider limits in one variable, say,
\[
\lim _{y \rightarrow q} f(x, y) \text { with } x \text { fixed.}
\]
If this limit exists for each choice of \(x\) from some set \(B \subseteq X,\) it defines a function
\[
g : B \rightarrow T
\]
with value
\[
g(x)=\lim _{y \rightarrow q} f(x, y), \quad x \in B.
\]
This means that
\[
(\forall x \in B)(\forall \varepsilon>0)(\exists \delta>0)\left(\forall y \in G_{\neg q}(\delta)\right) \quad \rho(g(x), f(x, y))<\varepsilon.
\]
Here, in general, \(\delta\) depends on both \(\varepsilon\) and \(x .\) However, in some cases (resembling uniform continuity), one and the same \(\delta\) (depending on \(\varepsilon\) only \()\) fits all choices of \(x\) from \(B .\) This suggests the following definition.
With the previous notation, suppose
\[
\lim _{y \rightarrow q} f(x, y)=g(x) \text { exists for each } x \in B(B \subseteq X).
\]
We say that this limit is uniform in \(x(\text {on } B),\) and we write
\[
"g(x)=\lim _{y \rightarrow q} f(x, y)(\text {uniformly for } x \in B),"
\]
iff for each \(\varepsilon>0,\) there is a \(\delta>0\) such that \(\rho(g(x), f(x, y))<\varepsilon\) for all \(x \in B\) and all \(y \in G_{\neg q}(\delta) .\) In symbols,
\[
(\forall \varepsilon>0)(\exists \delta>0)(\forall x \in B)\left(\forall y \in G_{\neg q}(\delta)\right) \quad \rho(g(x), f(x, y))<\varepsilon.
\]
Usually, the set \(B\) in formulas \((4)\) and \((5)\) is a deleted neighborhood of \(p\) in \(X,\) e.g.,
\[
B=G_{\neg p}(r), \text { or } B=X-\{p\}.
\]
Assume \((4)\) for such a \(B,\) so
\[
\lim _{y \rightarrow q} f(x, y)=g(x) \text { exists for each } x \in B.
\]
If, in addition,
\[
\lim _{x \rightarrow p} g(x)=s
\]
exists, we call s the iterated limit of \(f\) at \((p, q)\) (first in \(y,\) then in \(x ),\) denoted
\[
\lim _{x \rightarrow p} \lim _{y \rightarrow q} f(x, y).
\]
This limit is obtained by first letting \(y \rightarrow q\) (with \(x\) fixed \()\) and then letting \(x \rightarrow p .\) Quite similarly, we define
\[
\lim _{y \rightarrow q} \lim _{x \rightarrow p} f(x, y).
\]
In general, the two iterated limits (if they exist) are different, and their existence does not imply that of the double limit \((2),\) let alone \((3),\) nor does it imply the equality of all these limits. (See Problems 4ff below.) However, we have the following theorem.
(Osgood). Let \(\left(T, \rho^{\prime}\right)\) be complete. Assume the existence of the following limits of the function \(f : X \times Y \rightarrow T :\)
(i) \(\lim _{y \rightarrow q} f(x, y)=g(x)\) (uniformly for \(x \in X-\{p\} )\) and
(ii) \(\lim _{x \rightarrow p} f(x, y)=h(y)\) for \(y \in Y-\{q\}\).
Then the double limit and the two iterated limits of \(f\) at \((p, q)\) exist and all
three coincide.
- Proof
-
Let \(\varepsilon>0 .\) By our assumption (i), there is a \(\delta>0\) such that
\[
(\forall x \in X-\{p\})\left(\forall y \in G_{\neg q}(\delta)\right) \quad \rho(g(x), f(x, y))<\frac{\varepsilon}{4} \quad(\mathrm{cf} .(5)).
\]
Now take any \(y^{\prime}, y^{\prime \prime} \in G_{\neg q}(\delta) .\) By assumption (ii), there is an \(x^{\prime} \in X-\{p\}\) so close to \(p\) that
\[
\rho\left(h\left(y^{\prime}\right), f\left(x^{\prime}, y^{\prime}\right)\right)<\frac{\varepsilon}{4} \text { and } \rho\left(h\left(y^{\prime \prime}\right), f\left(x^{\prime}, y^{\prime \prime}\right)\right)<\frac{\varepsilon}{4} \cdot(\mathrm{Why} ?)
\]
Hence, using \(\left(5^{\prime}\right)\) and the triangle law (repeatedly), we obtain for such \(y^{\prime}, y^{\prime \prime}\)
\[
\begin{aligned} \rho\left(h\left(y^{\prime}\right), h\left(y^{\prime \prime}\right)\right) \leq & \rho\left(h\left(y^{\prime}\right), f\left(x^{\prime}, y^{\prime}\right)\right)+\rho\left(f\left(x^{\prime}, y^{\prime}\right), g\left(x^{\prime}\right)\right) \\ &+\rho\left(g\left(x^{\prime}\right), f\left(x^{\prime}, y^{\prime \prime}\right)\right)+\rho\left(f\left(x^{\prime}, y^{\prime \prime}\right), h\left(y^{\prime \prime}\right)\right) \\<& \frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4}=\varepsilon \end{aligned}
\]
It follows that the function \(h\) satisfies the Cauchy criterion of Theorem 2 in §2. (It does apply since \(T\) is complete.) Thus \(\lim _{y \rightarrow q} h(y)\) exists, and, by assumption ( ii), it equals \(\lim _{y \rightarrow q} \lim _{x \rightarrow p} f(x, y)\) (which therefore exists).
Let then \(H=\lim _{y \rightarrow q} h(y) .\) With \(\delta\) as above, fix some \(y_{0} \in G_{\neg q}(\delta)\) so close to \(q\) that
\[
\rho\left(h\left(y_{0}\right), H\right)<\frac{\varepsilon}{4}.
\]
Also, using assumption (ii), choose a \(\delta^{\prime}>0\left(\delta^{\prime} \leq \delta\right)\) such that
\[
\rho\left(h\left(y_{0}\right), f\left(x, y_{0}\right)\right)<\frac{\varepsilon}{4} \quad \text { for } x \in G_{\neg p}\left(\delta^{\prime}\right).
\]
Combining with \(\left(5^{\prime}\right),\) obtain \(\left(\forall x \in G_{-p}\left(\delta^{\prime}\right)\right)\)
\[
\rho(H, g(x)) \leq \rho\left(H, h\left(y_{0}\right)\right)+\rho\left(h\left(y_{0}\right), f\left(x, y_{0}\right)\right)+\rho\left(f\left(x, y_{0}\right), g(x)\right)<\frac{3 \varepsilon}{4}.
\]
Thus
\[
\left(\forall x \in G_{\neg p}\left(\delta^{\prime}\right)\right) \quad \rho(H, g(x))<\varepsilon.
\]
Hence \(\lim _{x \rightarrow p} g(x)=H,\) i.e., the second iterated limit, \(\lim _{x \rightarrow p} \lim _{y \rightarrow q} f(x, y),\) likewise exists and equals \(H .\)
Finally, with the same \(\delta^{\prime} \leq \delta,\) we combine \((6)\) and \(\left(5^{\prime}\right)\) to obtain
\[
\begin{array}{l}{\left(\forall x \in G_{\neg p}\left(\delta^{\prime}\right)\right)\left(\forall y \in G_{\neg q}\left(\delta^{\prime}\right)\right) \rho(H, f(x, y)) \leq \rho(H, g(x))+\rho(g(x), f(x, y))<\frac{3 \varepsilon}{4}+\frac{\varepsilon}{4}=\varepsilon}\end{array}
\]
Hence the double limit \((2)\) also exists and equals \(H . \square\)
Note 2.
The same proof works also with \(f\) restricted to\((X-\{p\}) \times(Y-\{q\})\) so that the "lines" \(x=p\) and \(y=q\) are excluded from \(D_{f} .\) In this case,
formulas \((2)\) and \((3)\) mean the same; i.e.,
\[
\lim _{x \rightarrow p \atop y \rightarrow q} f(x, y)=\lim _{(x, y) \rightarrow(p, q)} f(x, y).
\]
Note 3. In Theorem \(2,\) we may take \(E^{*}\) (suitably metrized) for \(X\) or \(Y\) or \(T .\) Then the theorem also applies to limits at \(\pm \infty,\) and infinite limits. We may also take \(X=Y=N \cup\{+\infty\}\) (the naturals together with \(+\infty ),\) with the same \(E^{*}\)-metric, and consider limits at \(p=+\infty .\) Moreover, by Note \(2,\) we may restrict \(f\) to \(N \times N,\) so that \(f : N \times N \rightarrow T\) becomes a double sequence (Chapter 1, §9). Writing \(m\) and \(n\) for \(x\) and \(y,\) and \(u_{m n}\) for \(f(x, y),\) we then obtain Osgood's theorem for double sequences (also called the Moore-Smith theorem) as follows.
Let \(\left\{u_{m n}\right\}\) be a double sequence in a complete space \(\left(T, \rho^{\prime}\right) .\) If
\[
\lim _{n \rightarrow \infty} u_{m n}=q_{m} \text { exists for each } m
\]
and if
\[
\lim _{m \rightarrow \infty} u_{m n}=p_{n}(\text {uniformly in } n) \text { likewise exists, }
\]
then the double limit and the two iterated limits of \(\left\{u_{m n}\right\}\) exist and
\[
\lim _{m \rightarrow \infty \atop n \rightarrow \infty} u_{m n}=\lim _{n \rightarrow \infty} \lim _{m \rightarrow \infty} u_{m n}=\lim _{m \rightarrow \infty} \lim _{n \rightarrow \infty} u_{m n}.
\]
Here the assumption that \(\lim _{m \rightarrow \infty} u_{m n}=p_{n}\) (uniformly in \(n )\) means, by \((5),\) that
\[
(\forall \varepsilon>0)(\exists k)(\forall n)(\forall m>k) \quad \rho\left(u_{m n}, p_{n}\right)<\varepsilon.
\]
Similarly, the statement " \(\lim _{m \rightarrow \infty \atop n \rightarrow \infty} u_{m n}=s^{\prime \prime}(\operatorname{see}(2))\) is tantamount to
\[
(\forall \varepsilon>0)(\exists k)(\forall m, n>k) \quad \rho\left(u_{m n}, s\right)<\varepsilon.
\]
Note 4.
Given any sequence \(\left\{x_{m}\right\} \subseteq(S, \rho),\) we may consider the double \(\operatorname{limit} \lim _{m \rightarrow \infty \atop n \rightarrow \infty} \rho\left(x_{m}, x_{n}\right)\) in \(E^{1} .\) By using \((8),\) one easily sees that
\[
\lim _{m \rightarrow \infty \atop n \rightarrow \infty} \rho\left(x_{m}, x_{n}\right)=0
\]
iff
\[
(\forall \varepsilon>0)(\exists k)(\forall m, n>k) \quad \rho\left(x_{m}, x_{n}\right)<\varepsilon,
\]
i.e., \(i f f\left\{x_{m}\right\}\) is a Cauchy sequence. Thus Cauchy sequences are those for which \(\lim _{n \rightarrow \infty \atop n \rightarrow \infty} \rho\left(x_{m}, x_{n}\right)=0\).
In every metric space \((S, \rho),\) the metric \(\rho :(S \times S) \rightarrow E^{1}\) is a continuous function on the product space \(S \times S\).
- Proof
-
Fix any \((p, q) \in S \times S .\) By Theorem 1 of §2, \rho\) is continuous at \((p, q)\) iff
\[
\rho\left(x_{m}, y_{m}\right) \rightarrow \rho(p, q) \text { whenever }\left(x_{m}, y_{m}\right) \rightarrow(p, q),
\]
i.e., whenever \(x_{m} \rightarrow p\) and \(y_{m} \rightarrow q .\) However, this follows by Theorem 4 in Chapter 3, §15. Thus continuity is proved. \(\square\)