Metric Spaces Past paper qestion

Asked Mar 16 '20 at 18:01

Active Mar 16 '20 at 18:33

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Let $(X, d)$ be a metric space. For each $a ∈ X$ let $f_a : X → \mathbb{R}$ be the function $f_a(x) = d(x, a)$.

(a) Prove that for all $a, b ∈ X$ $$\sup_{x∈X}|f_a(x) − f_b(x)| = d(a, b).$$

(b) Fix a point $c ∈ X$. For each point $a ∈ X$ define the function $g_a$ by $g_a(x) = f_a(x)−f_c(x)$. Show that for any $a ∈ X$ the function $g_a$ is bounded.

(c) Show that for any $a, b ∈ X$, $d_∞(g_a, g_b) = d(a, b)$ where $d_∞$ is the sup-metric on the space $B(X)$ of bounded functions on $X$.

Past paper question, but I don't have the solutions and don't know how to do it. I have done part (c). For a) how am I supposed to prove it if I don't know what the metric d is? I have $\sup_{x∈X}|f_a(x) − f_b(x)| = \sup_{x∈X}|d(x,a) − d(x,b)|$, but how can I say this is equal to $d(a,b)$ without the definition of d?

edited Mar 16 '20 at 18:33

asked Mar 16 '20 at 18:01

MathGeek1998

1

What have you tried? – user113102 Mar 16 '20 at 18:04
Just try the triangle inequality with $a,b,x$. It won't solve itself, you need to do some work. – copper.hat Mar 16 '20 at 18:11
How can I use the triangle inequality for (a)? or do you mean for part (b)? – MathGeek1998 Mar 16 '20 at 18:27
Can you show that it achieves $d(a,b)$? Then what does the triangle inequality say about a third point $c$. – user113102 Mar 16 '20 at 18:31
https://mathworld.wolfram.com/TriangleInequality.html – user113102 Mar 16 '20 at 18:34
The part I'm struggling with the most is how to show that it achieved d(a,b) in part (a), if you read the last part of the question I explained why – MathGeek1998 Mar 16 '20 at 18:34
What happens if you plug $a$ in, or $b$? – user113102 Mar 16 '20 at 18:35
For a) how am I supposed to prove it if I don't know what the metric $d$ is? I have $\sup_{x∈X}|fa(x)−fb(x)|=\sup_{x∈X}|d(x,a)−d(x,b)|$, but how can I say this is equal to $d(a,b)$ without the definition of $d$ or assuming that it is the euclidean metric? – MathGeek1998 Mar 16 '20 at 18:39
What is $d(a,a)$? – user113102 Mar 16 '20 at 18:40
$d(a,a)=0$ since d is a metric – MathGeek1998 Mar 16 '20 at 18:41
Ok so if I choose $x=a$, what is the value of the expression $f_a(x) - f_b(x)$? – user113102 Mar 16 '20 at 18:41
Then it is $-d(a,b)$ – MathGeek1998 Mar 16 '20 at 18:43
Great, but we have absolute values, so you've shown that $sup_{x \in X} | f_a(x) - f_b(x)|$ achieves the value $d(a,b)$. (at both $a$ and $b$). What remains to show is that for any other $c$ in $x$, you can bound it above by this value. – user113102 Mar 16 '20 at 18:44
Ok, but then I am saying that it is only true for $x=a$ right? Is this a valid proof? – MathGeek1998 Mar 16 '20 at 18:45
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You have shown that $d(a,b)$ is achieved, and so $d(a,b) \leq sup_{x\in X} |f_a(x) - f_b(x)|$. To show that it equals that supremum, you need to show that for any other value $c \in X$, $|f_a(c) - f_b(c)| \leq d(a,b)$. – user113102 Mar 16 '20 at 18:48
Thank you very much! – MathGeek1998 Mar 16 '20 at 18:54
Do you know how I should approach part (b) – MathGeek1998 Mar 16 '20 at 18:55
Think about how you might be able to use part (a) – user113102 Mar 16 '20 at 19:51

Metric Spaces Past paper qestion

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