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W.A. Sutherland's "Introduction to Metric and Topological Spaces" (1st edition, 1975, Oxford Science Publications) has this puzzling question in Chapter $2$: Continuity Generalized: Metric Spaces, Exercises $2.6: 17$:

"Given a metric space $M = \{A, d\}$, let $M \times M = \{A \times A, d_1\}$ be as in Example $2.2.7$. Show that $d: A \times A \to \mathbb R$ is a continuous mapping from $M \times M$ to $\mathbb R$."

In this context, example $2.2.7$ defines the product space:

"Given two metric spaces $M = \{A, d\}$, $M' = \{A', D'\}$, we define several metrics on $A \times A'$ ... If $x = (x_1, x_2)$ and $y = (y_1, y_2)$ are in $A \times A'$, let: $$d_1(x, y) = d (x_1, y_1) + d'(x_2, y_2)$$"

... which Sutherland defines as a the "taxicab metric".

The expression for what is presumed to be the object of the exercise seems to be wrong.

What is given $d: A \times A \to \mathbb R$ is merely the metric $d$ on $M$.

Whether $d$ is or is not continuous is independent of what $d_1$ may happen to be.

If the exercise were to prove that $d$ is continuous, then there would be no point in mentioning that it is a factor space of a Cartesian product.

Thus it is supposed that what we are expecting to prove is that the metric $d_1: (A \times A) \times (A \times A): \mathbb R$ is continuous.

As it happens, a metric is always a "continuous mapping" (that is, a "continuous real-valued function") no matter what $A$ or $d$ actually are.

Hence it is trivial to apply that result to $d_1$.

Is there anyone out there who may have an idea of what Sutherland was trying to get at here?

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    $\begingroup$ The goal of the exercise is to prove that $d$ viewed as a function from $A\times A$ to $\mathbb{R}$ is continuous with respect to the metric $d_1$ of $A\times A$. We will need to use certain properties of $d$ being a metric and the relation between $d$ and $d_1$; but we're only thinking about $d$ is a function, kinda forgetting that it's a metric on something. $\endgroup$ Commented Jan 9, 2021 at 20:51
  • $\begingroup$ @zipirovich Thanks for this, but I still don't really understand what it means. Are you able to craft an answer to this? I feel it would be instructive. $\endgroup$ Commented Jan 9, 2021 at 20:54

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Given metric spaces $(X,\rho_X)$ and $(Y,\rho_Y)$, and a function $f : X \to Y$, we say that $f$ is continuous with respect to $\rho_X$ and $\rho_Y$ (in shorthand, $f : (X,\rho_X) \to (Y,\rho_Y)$ is continuous) if for any $x_0 \in X$ and any $\varepsilon>0$ there exists $\delta>0$ (that depends on $x_0$ and $\varepsilon$) such that $$(\forall x \in X) \quad \rho_X(x,x_0)<\delta \!\implies\! \rho_Y(f(x),f(x_0)) < \varepsilon.$$ So, you need to prove that if $(A,d)$ is a metric space, then $d : (A \times A,d_1) \to (\mathbb R,|\cdot|)$ is continuous, that is, given $(x_0,y_0) \in A \times A$ and $\varepsilon>0$, you need to find $\delta>0$ such that $$(\forall x,y \in A) \quad d_1((x,y),(x_0,y_0)) < \delta \!\implies\! |d(x,y)-d(x_0,y_0)| < \varepsilon.$$

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  • $\begingroup$ Thank you, I understand what he is getting at now. Good job. $\endgroup$ Commented Jan 9, 2021 at 21:32
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Let $(x_0, y_0) \in A \times A$. If $d_1((x, y), (x_0, y_0)) = d(x, x_0) + d(y, y_0) < \epsilon$, then $$ |d(x, y) - d(x_0, y_0)| \le |d(x, y) - d(x, y_0)| + |d(x, y_0) - d(x_0, y_0)| \le 2\epsilon, $$ since in a general metric space it holds that if $d(x, y) < \epsilon$, then $d(z, x) - d(z, y) < \epsilon$; this is because $d(z, x) \le d(z, y) + d(x, y)$, and we may assume without loss of generality that $d(z, x) \ge d(z, y)$.

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