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I'm just curious why duality in mathematics is so common?

To be more concrete, here is an example: the harmonic mean as the dual of the mean. It looks almost like the opposite of the mean, yet it behaves quite similarly.

profPlum
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    I think your question is philosophical rather than mathematical so you may get a better response on https://philosophy.stackexchange.com. My naive comment would be that mathematics is full of deep correspondences between what seem at first sight to be unconnected subject areas, like complex analysis and number theory, or topology and group theory, or group theory and field theory. Those correspondences often work like mirror images between the two subjects, and those mirror images are formalised as dualities. – Rob Arthan Sep 21 '21 at 00:12
  • Perhaps some duals indicate connections between concepts that are not readily apparent. For example, a student new to Topology, when considering that $\left(A^c\right)^c = A$ might not imagine that in a Venn diagram, $A$ is colored blue, and $A^c$ is colored red, and that switching from a set to its complement analogizes to switching colors. That is, if the diagram only has two colors, and you switch colors twice (blue to red, then red to blue), you must return to your original color. Obvious to experienced people, perhaps not obvious to newbies. – user2661923 Sep 21 '21 at 00:14
  • @RobArthan I think that there is a lot of merit in your comment. However, I also think that there are purely mathematical aspects to the OP's question. It seems to me that often, someone will read and understand an analytical proof to a theorem, without fully grasping (at an instinctive level) why the theorem is true. This lack of understanding will lead to not immediately comprehending what might be regarded as a dual. Consider a Combinatorics problem, where induction is used. The solver has proven that the result is valid without direct analysis, and perhaps w/o knowing why it's true. – user2661923 Sep 21 '21 at 00:19
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    One of my favorite “dual” concepts is the Stirling Numbers of the second kind as duals of binomial coefficients. We can think if $\binom nk$ by counting the monomorphisms $[k]\to [n]$ modulo and equivalence class where $f\sim g$ if $f\circ \pi=g$ for some isomorphism $\pi:[k]\to [k].$ The Stirling case is the dual of this, counting equivalence classes of epimorphism $[n]\to[k]$ modulo again isomorphisms $[k]\to[k].$ These two concepts are duals, but they are very different. – Thomas Andrews Sep 21 '21 at 00:39

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