29 Sep 2015

to \aleph 1 and beyond

Today’s lecture was on code abstraction. In one part Professor Olivier Danvy gave an example on how recursion can be viewed from a more mathematical point of view through demonstrating the \(\mathbb{N}\)-bijective nature of some set of procedures. Here is the idea:

as more procedures are enumerated in this a manner, we would be able to do factorial on a larger subset of the natural numbers. There isn’t really recursion going on since there is no self-referencing. And this can continue to infinity. But as anyone with a common sense would know, this goes as far as \(\aleph_0\) by virtue to the fact that it is bijective to \(\mathbb{N}\).

So a curious question arises: let \(F_{\mathbb{N}}\) be some set of procedures that can be enumerated like above, is it possible to construct a set of procedures, \(F_{\mathbb{R}}\), similar to \(F_{\mathbb{N}}\) (the notation of it being similar to \(F_{\mathbb{N}}\) is rather tricky here: let’s just say it is similar in that it is related to recursion, or that it is a generalization of \(F_{\mathbb{N}}\)), that has a cardinality equal or greater than \(\aleph_1\)? Since such set of procedures cannot be enumerated as George Cantor had demonstrated more than a century ago (O! the whirligig of time), it would need to be constructed in a different manner.

The same question can be asked regarding Turing machine: can we formulate a model of computation, similar to that of Turing’s (or more of a generalization of it), such that there exists a single object, \(U\), that can (similar to how every Turing machine can be simulated by a Universal Turing machine) simulate every object in this model, wherein every \(r \in \mathbb{R}\) can be used to encode a distinct object? Perhaps in such model, there exists some object that can solve a subset of the Halting problem previously unsolvable (i.e. those which halt in a set of steps with a cardinality less than \(\aleph_1\) (and now we have successfully drawn the continuum hypothesis into the discussion)).

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