Let $(X,d)$ be a metric space. Take $\mathbb{R}$ with its Euclidean metric, $e$ , and $X \times X$ with one of the canonical metrics on the Cartesian product. Prove that $q : X \times X \rightarrow \mathbb{R}^+ , (u,v) \mapsto d(u, v)$ is continuous. Given metric spaces, $(X,d)$ and $(X_0,d_0)$ , the function $f : X\rightarrow Y$ is uniformly continuous if and only if given any $\varepsilon > 0$ there is a $\delta > 0$ with $\mathbb{R}$ $0$ whenever \( d(u,v)

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Algebra 1

Dilations of functions

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Q. Let

(X,d)

be a metric space. Take

\mathbb{R}

with its Euclidean metric,

e

, and

X \times X

with one of the canonical metrics on the Cartesian product. Prove that

q : X \times X \rightarrow \mathbb{R}^+ , (u,v) \mapsto d(u, v)

is continuous. Given metric spaces,

(X,d)

and

(X_0,d_0)

, the function

f : X\rightarrow Y

is uniformly continuous if and only if given any

\varepsilon > 0

there is a

\delta > 0

with

\mathbb{R}

0

whenever \( d(u,v)

Define $\epsilon$ and $\delta$ : We need to show that for any $\epsilon > 0$ , there exists a $\delta > 0$ such that for all $(u,v), (u',v')$ in $X \times X$ , if the distance between $(u,v)$ and $(u',v')$ is less than $\delta$ , then the absolute difference between $d(u, v)$ and $\delta$ $0$ is less than $\epsilon$ .
Apply triangle inequality: Let's choose $\varepsilon > 0$ . We want to find $\delta > 0$ such that if $d_{X\times X}((u,v), (u',v')) < \delta$ , then $|d(u, v) - d(u', v')| < \varepsilon$ , where $d_{X\times X}$ is the metric on $X \times X$ .
Use max metric: We can use the triangle inequality property of the metric $d$ on $X$ , which states that for any $x, y, z$ in $X$ , $d(x, z) \leq d(x, y) + d(y, z)$ .
Choose $\delta = \frac{\epsilon}{2}$ : Applying the triangle inequality twice, we get $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ and $d(u', v') \leq d(u', u) + d(u, v) + d(v, v')$ .
Choose $\delta = \frac{\epsilon}{2}$ : Applying the triangle inequality twice, we get $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ and $d(u', v') \leq d(u', u) + d(u, v) + d(v, v')$ . Subtracting the second inequality from the first, we obtain $|d(u, v) - d(u', v')| \leq d(u, u') + d(v, v')$ .
Choose $\delta = \frac{\epsilon}{2}$ : Applying the triangle inequality twice, we get $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ and $d(u', v') \leq d(u', u) + d(u, v) + d(v, v')$ . Subtracting the second inequality from the first, we obtain |d(u, v) - d(u', v')| \leq d(u, u') + d(v, v')\. Now, we need to relate \$d(u, u') and $d(v, v')$ to the metric $d_{X\times X}$ on $X \times X$ . If we choose the metric $d_{X\times X}$ to be the max metric, for example, then $d_{X\times X}((u,v), (u',v')) = \max\{d(u, u'), d(v, v')\}$ .
Choose $\delta = \frac{\epsilon}{2}$ : Applying the triangle inequality twice, we get $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ and $d(u', v') \leq d(u', u) + d(u, v) + d(v, v')$ . Subtracting the second inequality from the first, we obtain |d(u, v) - d(u', v')| \leq d(u, u') + d(v, v')\. Now, we need to relate \$d(u, u') and $d(v, v')$ to the metric $d_{X\times X}$ on $X \times X$ . If we choose the metric $d_{X\times X}$ to be the max metric, for example, then $d_{X\times X}((u,v), (u',v')) = \max\{d(u, u'), d(v, v')\}$ . If $d_{X\times X}((u,v), (u',v')) < \delta$ , then both $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $0$ and $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $1$ . We can choose $\delta = \frac{\epsilon}{2}$ to ensure that $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $3$ .
Choose $\delta = \frac{\epsilon}{2}$ : Applying the triangle inequality twice, we get $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ and $d(u', v') \leq d(u', u) + d(u, v) + d(v, v')$ . Subtracting the second inequality from the first, we obtain $|d(u, v) - d(u', v')| \leq d(u, u') + d(v, v')$ . Now, we need to relate $d(u, u')$ and $d(v, v')$ to the metric $d_{X\times X}$ on $X \times X$ . If we choose the metric $d_{X\times X}$ to be the max metric, for example, then $d_{X\times X}((u,v), (u',v')) = \max\{d(u, u'), d(v, v')\}$ . If $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $0$ , then both $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $1$ and $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $2$ . We can choose $\delta = \frac{\epsilon}{2}$ to ensure that $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $4$ . Therefore, for any $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $5$ , we can find a $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $6$ (specifically $\delta = \frac{\epsilon}{2}$ ) such that if $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $0$ , then $d(u, v) \leq d(u, u') + d(u', v') + d(v', v)$ $9$ , which proves that q is continuous.