active draft. a technical sketch. general, before special —alignment, before distraction

the improbable yet elementary case

1. paradigm, measure, common measures, incommensurability

$Pa \;,\; Me \;,\; {}_{Me}^{\cap} \;,\;{}_{Me}^{\cap\varnothing}$

A reinterpretation of thomas kuhn’s ‘on the structure of scientific revolutions’, through the lens of set-theory(-ish) mathematics ¹.

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note: consider all set-theory as pseudo-set-theory; a means for a novice mathematician to express ideas in less time and fewer words, than a similarly novice writer might, in prose.

$Pa$ , $Me$

Let us consider a paradigm $Pa$ , as a set of two measures $Me_{1}$ and $Me_{2}$ : $$Pa = \{ Me_{1}, Me_{2}\} : |Pa| = 2$$ $$\ldots$$

$\cup$ , $\cap$

If paradigm $Pa_{1}$ , contains measures $Me_{1, 2, 3}$ : $$Pa_{1} = \{ {Me_{1}, Me_{2}, Me_{3}}\}$$ And paradigm $Pa_{2}$ , contains measures $Me_{2, 3, 4}$ : $$Pa_{2} = \{{Me_{2}, Me_{3}, Me_{4}}\}$$ $$\ldots$$ The set-of-all measures ${}^{\cup}_{Me}$ , across $Pa_{1}$ and $Pa_{2}$ , can be found by union $\cup$ : $${}_{Me}^{\cup} = Pa_{1} \cup Pa_{2} = \{ Me_{1}, Me_{2}, Me_{3}, Me_{4}\}$$ $$\ldots$$ The set-of-common measures ${}^{\cap}_{Me}$ , between $Pa_{1}$ and $Pa_{2}$ , can be found by intersection $\cap$ : $${}^{\cap}_{Me} = Pa_{1} \cap Pa_{2} = \{ Me_{2}, Me_{3}\}$$ $$\ldots$$ Observing: $$|{}_{Me}^{\cup}| = 4 \;,\; |{}^{\cap}_{Me}| = 2 \;,\;|{}^{\cap}_{Me}| \lt |{}^{\cup}_{Me}|$$ $$\ldots$$

note: remember, this is a simplification, and an introduction

$\cap\varnothing$

Considering paradigm $Pa_{3}$ , and paradigm $Pa_{4}$ , whereby : $$Pa_{3} = \{ {Me_{1}, Me_{2}, Me_{3}}\}$$ $$Pa_{4} = \{ {Me_{4}, Me_{5}, Me_{6}}\}$$ $$\ldots$$ When paradigms $Pa_{3}$ and $Pa_{4}$ , do not share common measures, then ${}^{\cap}_{Me}$ , is an empty set $\varnothing$ : $${}^{\cap}_{Me} = Pa_{3} \cap Pa_{4} = \varnothing : |\varnothing| = 0$$ And paradigms $Pa_{3}$ and $Pa_{4}$ , can be said to be incommensurable ${}^{\cap\varnothing}_{Me}$ : $${}^{\cap}_{Me} = Pa_{3} \cap Pa_{4} = \varnothing : {}^{\cap}_{Me} \rightarrow {}^{\cap\varnothing}_{Me} \;, \; |{}^{\cap\varnothing}_{Me}| = 0$$ $$\ldots$$

note: while any two paradigms may appear incommensurable as an isolated pair ², we will later discover that there exists a universally special paradigm, which by analysis or composition, renders all paradigms reconcilable, and as such, commensurable

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