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Re: Disjoint, subset, & interior, or more...???



Nate Hayes wrote:
Arnold Neumaier wrote:
Nate Hayes wrote:

According to Motion 3, and in accordance with standard mathematical
practice, [1,Inf] is unbounded, no matter how decorations handle this.

If we cannot rely on accepted motions to discuss further motion
we'll always remain on ground zero and never get anywhere.


For example, on the one hand Arnold argues that the "interior" relation in Motion 13.04 is not topological interior. THis criticism is valid only for unbounded intervals. On the other hand, Arnold also advocates an IsBounded deocration. In that case, the definition for "interior" in Motion 13.04 _is_ the correct definition of topological interior (by his own logic and reasoning as shown in recent e-mails in this forum).

No. No matter how intervals are represented, the inequality x>=1 always
defines the unbounded interval [1,Inf], and not a bounded surrogate
[1,Overflow] without a meaning as a set of real numbers.

This contradicts your previous statements and postings that
   [1,Infinity] \subseteq [1,Infinity]
should be "false" if both intervals are adorned with the IsBounded decoration.

No, since x>=1 translates into (1,Inf],isUnbounded).