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Mr Example.

Name: Anonymous 2007-08-30 13:43 ID:eUFPtE+p

I enjoyed those two Monsieur Ejemplé threads so I thought I'd pose a maths question of my own, in fact I'll do two on countability. They're not that hard, but I think they're more enjoyable than calculus

1. Let f : R -> R be monotonic.
Is  the set { x | f is discontinous at x} countable?
where x is in R.


2. A function f : N -> N is increaing if f(n)>= f(n+1)   (if it's bigger than OR equal to) and a decreasing function is similarly defined.

is the set {f | f is increasing} countable?
is the set {f | f is decreasing} countable?

Name: Anonymous 2007-09-02 7:18 ID:zsqHUXtC

It only generates finite binary sequences.

I had a good think about it last night, and the binary sequences are represented by figures of eights.

An infinite sequence does not correspond to a figure of eight, because it's an infinite sequence.

The only sequences that can be represented are finite, I'm pretty certain.


Anyway my proof works because there are two seperate areas.

A figure of eight can share an area with another shape, but it must either fully enclose, or be fully enclosed within one of the loops.

Whereas I am assigning them a pair of points, on inside EACH loop. No two figures of eight can have the same pair and points, that fact it obvious.

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