33
$\begingroup$

Given a metric space $(X,d)$, how to prove that the function $d \colon X \times X \to \mathbf{R}$ is continuous?

If we take any two arbitrary real numbers $a$ and $b$ such that $a < b$, then we need to show that the set $d^{-1} (a,b)$ given by

$$ d^{-1} (a,b) := \{ (x, y) \in X \times X | a < d(x,y) < b \} $$

is open in the product topology on $X \times X$.

A basis for this product topology is the collection of all cartesian products of open balls in $(X,d)$.

$\endgroup$
1

2 Answers 2

28
$\begingroup$

For $a, b ∈ \mathbb{ℝ}$ such that $a < b$, let the ordered pair $x \times y ∈ d^{-1} \big( (a..b) \big)$, that is, let $x \times y \in X \times X$ such that $$ a < d(x,y) < b. \tag{0} $$ Now choose a real number $ε$ such that $U_{2ε} (d(x,y)) \subset (a..b)$, that is, choose $\epsilon$ such that $$ 0 < \epsilon < \frac{1}{2} \min \big\{ d(x, y) -a, \, b - d(x, y) \big\}. \tag{1} $$ Now we show that $$ B_d (x ; \epsilon ) \times B_d( y ; \epsilon) \subset d^{-1} \big( (a, b) \big), \tag{2} $$ where $$ B_d(x; \epsilon ) := \{ p \in X \colon d(p, x) < \epsilon \}. \tag{Definition 0} $$

For any ordered pair $x' \times y' ∈ B_d (x ; \epsilon ) \times B_d( y ; \epsilon)$, you have $$ d\left(x',y' \right) ≤ d(x',x) + d(x,y) + d(y,y') < d(x,y) + 2ε < b. \tag{3} $$ Also $$ d(x,y) ≤ d(x,x') + d(x',y') + d(y',y) < d(x',y') + 2ε, $$ which implies $$ a < d(x, y) - 2ε < d(x',y'). \tag{4} $$ From (3) and (4), we get $$ a < d(x,y) - 2ε < d(x',y') < d(x,y) + 2ε < b, $$ that is, $$ d(x',y') \in (a, b). $$ Therefore (2) above holds.

Thus we have shown that, for any point $x \times x \in X \times X$ such that $x \times y \in d^{-1} \big( (a, b) \big)$, there exists a basic open set $B := B_d (x ; \epsilon ) \times B_d( y ; \epsilon)$ for the product topology on $X \times X$ induced by the $d$-metric topologies on both the factors $X$ such that $$ x \times y \in B \subset d^{-1} \big( (a, b) \big). $$ Hence $d^{-1} \big( (a, b) \big)$ is an open set in the product space $X \times X$.

Thus for every open interval $(a, b)$ on the real line $\mathbb{R}$ (with the usual topology) the inverse image set $d^{-1} \big( (a, b) \big)$ is also an open set in the product space $X \times X$. So the function $d \colon X \times X \longrightarrow \mathbb{R}$ is continuous.

$\endgroup$
4
  • 1
    $\begingroup$ Sorry, I didn't recognize the notation, but by $(a..b)$ do you mean the set of all real numbers from $a$ to $b$ exclusive (i.e. the open interval)? $\endgroup$ Commented Jul 16, 2016 at 21:07
  • 4
    $\begingroup$ Yeah, it’s a notation introduced by Knuth, I believe. It’s quite handy. It avoids the confusion of regarding $(a,b)$ both as a tuple and as an open interval. See the good notations thread on mathoverflow. $\endgroup$
    – k.stm
    Commented Jul 17, 2016 at 17:53
  • 1
    $\begingroup$ So in the metric space $X$, we can take any point $x, y$ such that $d(x,y)=k$ with any given $k \in R$ ? $\endgroup$
    – RopuToran
    Commented Jan 2, 2021 at 17:21
  • $\begingroup$ @RopuToran No, why would you think that? In ℚ, you find no two points with distance exactly $\sqrt 2$. $\endgroup$
    – k.stm
    Commented Jan 3, 2021 at 13:28
4
$\begingroup$

Let $(x,y)\in d^{-1}(a,b)$. Define $\epsilon=\frac{1}{100}\min(d(x,y)-a,b-d(x,y))$. Then for any point $(\xi,\gamma)\in B(x,\epsilon)\times B(y,\epsilon)$, we have $${\color{blue} d(\xi,\gamma) }\leq d(\xi,x)+d(x,y)+d(y,\gamma)<2\epsilon+d(x,y)<{\color{blue}b}$$ and $${\color{blue} a}<d(x,y)-2\epsilon <d(\xi,\gamma)+\underbrace{ d(x,\xi)-\epsilon}_{<0} +\underbrace{d(y,\gamma)-\epsilon}_{<0}<{\color{blue} d(\xi,\gamma)}.$$ That is to say $$(x,y)\in B(x,\epsilon)\times B(y,\epsilon)\subseteq d^{-1}(a,b).$$

Hence any point $(x,y)\in d^{-1}(a,b)\subseteq X\times X$ is surrounded by some open set contained in $d^{-1}(a,b)$.

(After typing this answer, I realized this is EXACTLY the same as the answer by @k.stm; so thanks also to his contribution. I guess it will be hard to write a different one.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .