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I'm trying to prove this triangle inequality where $A \triangle B = A \cup B - A \cap B$ $$ \frac{|A \triangle B|}{|A| + |B|} \leq \frac{|A \triangle C|}{|A| + |C|} + \frac{|B \triangle C|}{|B| + |C|} $$

I came up with this and it doesn't work. $$ \frac{|A \triangle C|}{|A| + |C|} + \frac{|B \triangle C|}{|B| + |C|} \geq \frac{|A \triangle C| + |B \triangle C|}{|A \cap B| + |C|} \geq \frac{|A \triangle C \cup B \triangle C|}{|A \cap B| + |C|} $$

Any help would be appreaciated.

I just found it doesn't satisfy the triangle inequality, for example when A={0}, B={1}, C={0, 1}, 1 > 1/3 + 1/3. This post can be closed.

Harrison
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    How is defined $\frac{X}{Y}$ among sets? – Raffaele Jan 23 '21 at 16:44
  • @Raffaele: I guess that is set difference and $\le$ might be subset relation. But I donot have a guess for "+". – Bumblebee Jan 23 '21 at 17:15
  • @Bumblebee I knew that. It's the "fraction" that is unclear – Raffaele Jan 23 '21 at 17:17
  • @Raffaele: Agree with you. OP should carefully introduce his notations. – Bumblebee Jan 23 '21 at 17:20
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    Maybe $\frac{A}{B}:=\frac{|A|}{|B|}$ where $|S|$ is the cardinal (number of elements) of finite set $S$... – Jean Marie Jan 23 '21 at 18:01
  • Is $A+B:=|A|+|B|$ ? – Jean Marie Jan 23 '21 at 18:17
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    sorry just got back. @JeanMarie you are right. $\frac{X}{Y}$ is the number of items in X divided by the number of items in Y. I will correct my notations. – Harrison Jan 24 '21 at 00:00
  • @JeanMarie |△| / ||+|| = (1 + 1) / (1 + 1) = 1 – Harrison Jan 24 '21 at 11:19
  • Indeed, you are right. I erase my wrong calculation... Personally, I had been working on this kind of "pseudo-distances" linked to "Jaccard index". See this question: https://math.stackexchange.com/q/2119199. You will find there as well a reference to a question of mine. – Jean Marie Jan 24 '21 at 12:20
  • Thank you @JeanMarie, speaking of Jaccard index like metrics, do you happen to know any similar metrics which will have different distances even when 2 sets are not overlapping? For example, the Jaccard index of 2 non-overlapping spans are always 0, so it cannot measure how far apart the 2 spans are, i.e. {0, 1, 2} and {3,4} are as bad as {0, 1, 2} and {99,100}. – Harrison Jan 25 '21 at 02:10
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    As I understand it, what you are looking for would be a "mixture" of 2 "metrics", one purely based on sets and another one inherited from the usual distance on $\mathbb{N}$ for example. Have you heard about Hausdorff distance? it is a very effective distance between sets in a metric space. – Jean Marie Jan 25 '21 at 09:28
  • @JeanMarie Yes, I tried the Hausdorff distance which is the gap between 2 spans in the span metric space. I also tried to use a distance on the length of the spans, i.e. ||A| - |B||, however their scales are much larger than the Jaccard distance. I have to normalise them first. Are there any conventional ways to standardise these distances? – Harrison Jan 25 '21 at 10:35
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    There are different "standardisations" of distances, like the ones you can see in this question – Jean Marie Jan 25 '21 at 11:03
  • Having a distance $\le 1$ has some merits, in particular "psychological" (scale $0-100 %$) and/or with connection with probabilities' range. – Jean Marie Jan 25 '21 at 11:31

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