• saigot ( saigot@lemmy.ca ) English
    64·
    1 year ago

    It’s been a while but here we go:

    for orange to be a metric 4 conditions must be met:

    1. 🍊(🍎,🍎) = 0
    proof

    since 🍎(x) - 🍎(x) will always be 0 for any 🍎 and any x in domain

    1. 🍊(🍎,🍌) > 0 if 🍎 != 🍌.
    proof

    |🍎(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 -> 🍎=🍌

    1. 🍊(🍎,🍌) = 🍊(🍌,🍎)
    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 🍊(🍎,🍌) = 🍊(🍌,🍎)

    1. The Triangle Inequality:🍊(🍎,🍇) <= 🍊(🍎,🍌) + 🍊(🍌, 🍇)
    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!

    • OmnipotentEntity ( OmnipotentEntity@beehaw.org ) English
      21·
      1 year ago

      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.

    • subiprime ( subiprime@lemmy.blahaj.zone ) English
      2·
      1 year ago

      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

      • saigot ( saigot@lemmy.ca ) English
        3·
        1 year ago

        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

    • OmnipotentEntity ( OmnipotentEntity@beehaw.org ) English
      12·
      9 months ago

      Anyway, to prove this is a metric we must prove that it satisfies the 4 laws of metrics.

      1. The distance from a point to itself is zero. 🍊 (🍎, 🍎) = 0

      This can be accomplished by simply observing that |🍎 (x) - 🍎 (x)| = 0 ∀x ∈ [a,b], so its sup = 0.

      2. The distance between any two distinct points is non-negative.

      If 🍎 ≠ 🍌, then ∃x ∈ [a,b] such that 🍎 (x) ≠ 🍌 (x). Thus for this point |🍎 (x) - 🍌 (x)| > 0 and the sup > 0.

      3. 🍊 (🍎, 🍌) = 🍊 (🍌, 🍎) ∀(🍎, 🍌) in our space of functions.

      Again, we must simply apply the definition of 🍊 observing that ∀x ∈ [a,b] |🍎 (x) - 🍌 (x)| = |🍌 (x) - 🍎 (x)|, and the sup of two equal sets is equal.

      4. Triangle inequality, for any triple of functions (🍎, 🍌, 🍇), 🍊 (🍎, 🍌) + 🍊 (🍌, 🍇) ≥ 🍊 (🍎, 🍇)

      For any (🐁, 🐈, 🐕) ∈ ℝ³ it is well known that |🐁 - 🐕| ≤ |🐁 - 🐈| + |🐈 - 🐕|, (triangle inequality of absolute values).

      Further, for any two nonnegative functions 🍍, 🍑 we have sup({🍍 (x) : x ∈ [a, b]}) + sup({🍑 (x) : x ∈ [a, b]}) ≥ sup({🍍 (x) + 🍑 (x) : x ∈ [a, b]})

      Letting 🍍 (x) = |🍎 (x) - 🍌 (x)|, and 🍑 (x) = |🍌 (x) - 🍇 (x)|, we have the following chain of implications:

      🍊 (🍎, 🍌) + 🍊 (🍌, 🍇) = sup(🍍 (x) : x ∈ [a, b]}) + sup({🍑 (x) : x ∈ [a, b]}) ≥ sup({🍍 (x) + 🍑 (x) : x ∈ [a, b]}) ≥ sup({🍎 (x) - 🍇 (x)| : x ∈ [a, b]) = 🍊 (🍎, 🍇)

      Taking the far left and far right side of this chain we have our triangles inequality that we seek.

      Because 🍊 satisfies all four requirements it is a metric. QED.

      QED stands for 👸⚡💎, naturally