Discussion about this post

User's avatar
Mark Young's avatar

(Quoting Florence Bacus)

> “One cannot rationally judge that an agent ought to do something they judge they cannot do.”

I think in this case the intent was that the first "they" refers back to "one" while the second "they" refers back to "an agent". That, at least, is the interpretation I came up with when trying to make it come out true. That is:

== J cannot rationally judge that R ought to do something J judges R unable to do.

> Also not necessarily true. Randy believes he is possessed by the devil and must burn my house to the ground; he judges that he cannot do otherwise. This is in fact false. But he still ought not to burn my house down!

Given my interpretation of the claim, your example here does not refute it. Randy may judge that he cannot do otherwise, but *you* judge that he can ("Randy perfectly is capable of staying home and cowering"). Thus it is OK (rationally) for *you* to judge that he ought not burn down your house.

> So what is it about my assertion that is “irrational”?

Nothing.

Expand full comment
Mark Young's avatar

I believe pen-paper-continuity (PPC) can be defined in terms of epsilon-delta continuity (EDC).

Consider making a drawing a shape without lifting your pen from the paper. Let f: T --> P be the function describing the pen's position of the paper at all times from when you start drawing to when you finish. (T is the time interval spent drawing the shape; P is the set of all points on the paper that the drawing is on.)

(For those who just barely followed the epsilon-delta definition, in this case |f(y) - f(c)| refers to the distance between f(y) and f(c), since those are points on a piece of paper (R^2) instead of points on a line (R).)

PPC = for every open interval I in T, the restriction of f to I is EDC at every point in I.

If at any time t during the drawing process you lifted the pen off the paper there will be an open interval with that time as its lower bound during which f(t) is undefined and so f is not EDC at t and so not PPC at any interval containing t. Conversely, if f is EDC at every point in an open interval I then it's defined at every point in that interval and you didn't lift your pen off the paper during that interval (no infinitely fast jumps of the pen).

Thomae’s function is not PPC anywhere, even tho it is EDC at uncountably many points. There is no open interval in which all points are EDC, because any open interval contains at least one rational number, at which the function is not EDC.

Anyone see a problem with that?

Expand full comment
3 more comments...

No posts