this post was submitted on 12 Apr 2025
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It's been a while but here we go:
for orange to be a metric 4 conditions must be met:
proof
since 🍎(x) - 🍎(x) will always be 0 for any 🍎 and any x in domainproof
|🍎(x) - 🍌(x)| >= 0 by definition, so 🍊(🍎,🍌) must be >= 0. we only have to prove that:
🍊(🍎,🍌) = 0 -> 🍎=🍌
Consider the contrapositive: 🍎!=🍌 -> 🍊(🍎,🍌) != 0
since 🍎!=🍌 ∃x s.t 🍎(x) != 🍌(x)
but then |🍎(x) - 🍌(x)| > 0
thus 🍊(🍎,🍌) > 0
thus 🍊(🍎,🍌) = 0 -> 🍎=🍌
proof
|🍎(x) - 🍌(x)| = |-1(-🍎(x) + 🍌(x))||-1(-🍎(x) + 🍌(x))| = |-1(🍌(x) - 🍎(x))|
|-1(🍌(x) - 🍎(x))| = |🍌(x) - 🍎(x)| since |-q| =|q|
so for any x |🍎(x) - 🍌(x)| = |🍌(x) - 🍎(x)|
which means 🍊(🍎,🍌) = 🍊(🍌,🍎)
proof
let x be the element in [a,b] s.t |🍎(x) - 🍇(x)| is maximized
let y be the element in [a,b] s.t |🍎(y) - 🍌(y)| is maximized
let z be the element in [a,b] s.t |🍌(z) - 🍇(z)| is maximized
🍊(🍎,🍇) <=🍊(🍎,🍌) + 🍊(🍌, 🍇) is equivalent to
|🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) - 🍇(x)|
Let's start with the following (obvious) inequality:
|🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)|
|🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| since |🍎(y) - 🍌(y)| is maximized
|🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| >= |🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)| since |🍌(z) - 🍇(z)| is maximized
|🍎(x) -🍌(x)| +|🍌(z) - 🍇(z)| >= ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| since |q| + |p| >= 0 so |q| + |p| = ||q| +|p||
||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| since |q| >= q forall q
therefore |🍎(y) -🍌(y)| +|🍌(z) - 🍇(z)| >= |🍎(x) - 🍇(x)|
since all 4 conditions are satisfied the 🍊 is a metric!
I don't understand any of this, but I upvoted because you showed your work
Here is my attempt to eli5, a metric is a formalized/generalized way to describe distance. Smart people thought about what makes distance distance and basically made a set of rules. Distance is a function where the distance between a point and itself is 0 (and only 0 in that case), is always positive, is the same distance whether you are coming or going and that going to a place and then another place has at least as much distance as just going to the last place (which is kind of the same as saying the shortest path between 2 points is a straight line).
You can see how these rules apply to point in 3d(or 2d) space and our intuitive understanding of distance between them. For example If a store is 2km going to a bank then the store is at least 2km but maybe more and if its 2km from home to the store its also 2km from the store to home. This might seem obvious, and it is for 3d space, but we can take it and apply it to all kinds of things.
This question is intentionally convoluted, but one way of conceptualizing it is: 🍎🍇🍌 are each functions that takes one value and spits out another. If you would graph this function it makes a line. 🍊 takes 2 lines and tells us how far apart they are, you can think about many ways to compare how far apart 2 line are, but the one given to us is to just take the one where the difference between the heights of the lines is greatest. For an example lets say 🍎 is the price of eggs and 🍇 is the price of organic eggs then 🍊(🍎,🍇) would give us the biggest difference in price there has ever been between them.
Our task in the problem is to prove that that idea of distance given to us follows the same rules as our intuitive definition of distance.
E: I originally misread the values the functions took as 2 dimensional coordinates, but it is really just 1 dimensional data, so I changed the metaphor.
Careful ⚠️ there is not guaranteed to be an element such that |🍎(x) - 🍇(x)| is maximized. Consider 🍎 (x) = x if x < 3, 0 otherwise. Let 🍇 (x) = 0, and let the domain be [0, 4]. Clearly, the sup(|🍎 (x) - 🍇 (x)| : x ∈ [0, 4]) = 3, but there is no concrete value of x that will return this result. If you wish to demonstrate this in this manner, you will need to introduce an 🐘 > 0 and do some pedantic limit work.
That is a fair criticism that I am too lazy to work out the details of 😊.
However, 🚨 note that the interval is closed and bounded and 🍎 and 🍇 are continuous (your 🍎 isn't), so by the EVT the maximum is obtained (but might not be unique).
Oh! My bad! I completely missed that the functions were continuous (it isn't required for 🍊 to be a metric)
Uh
Now do it for feet and miles
you mean something like this?
🦶🏿(🕷️,🕷️) = 0
🦶🏿(🕷️,☢️) > 0 if 🕷️ != ☢️.
🦶🏿(🕷️,☢️) = 🦶🏿(☢️,🕷️)
The Triangle Inequality:🦶🏿(🕷️,⚡) <= 🦶🏿(🕷️,☢️) + 🦶🏿(☢️, ⚡)
Now it makes sense. Thank you. I always thought foot spaces were more intuitive
I'm confused about this step in the final condition's proof:
|🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| since |q| >= q forall q
I can see how it's true by proving that |p| + |q| >= |p + q|, but that's not stated anywhere and I can't figure out how |q| >= q forall q is relevant.
Also, thanks a lot for making/showing a proof :D
It should be ||🍎(x) -🍌(x)| +|🍌(x) - 🍇(x)|| >=|🍎(x) -🍌(x) +🍌(x) - 🍇(x)| = |🍎(x) - 🍇(x)| I missed the abs that I added in the previous step.
let me make the variables less annoying:
||x-y|+|y-z|| >= |x-y+y-z| = |x-z| we are getting rid of the abs around |x-y| and |y-z| so the 2 y's can cancel out. We can do this because |x-y| >= x-y because |q| >= q
I think this could use a bit more elaboration, since if x-y+y-z < -(|x-y|+|y-z|), then ||x-y|+|y-z|| >= |x-y+y-z| wouldnt be true. This is impossible though since q >= -|q|
I see, thanks! :3