prove that no closed interval is the union of two mutually exclusive closed poin

Question

prove that no closed interval is the union of two mutually exclusive closed point sets.

Notes:

you could use the Completness Axion __> if M is a point set and there is a point to the right of every point of M then there is either a right-most point of M or a first point to the right of M.

also you could use if f is a continuous function whose domain includes the closed interval [a,b] and there is a point x in [a,b] so that f(x) is greater than or equal to zero then the set of all numbers x element of [a,b] such that f(x) >= 0 is a closed point set.

you could use this as well if f is a continuouse function whose domain includes a closed interval [a,b] and p is an element of [a,b] then the set of all numbers x is an elemnt [a,b] such that f(x) = f(p) is a closed point set.

you also could use the definition that says the statment that the point sets H and K are disjointed or mutually exclusive means that they have no point in common.

Explanation / Answer

Proof:
Let H=[aH,bH] and K=[aK,bK] where H and K are mutually exclusive.
Since H and K are both closed point sets they each have a leftmost point and a right most point (by the definition of a closed point set).
If there is a point pH between any two points m and n of K, then by axiom 1.6, there is a point q between both (m and q) and (q and n). Thus HK is not a closed interval.
If H<K then by axiom 1.6 there exists some point pxbetween the right most point of H
and the leftmost point of K where pxHK and again, no closed interval exists.
By symmetrical argument for the case where K<H there is a point between the leftmost point and right most point of KH that is not a member.
the union of two mutually exclusive closed point sets in not a closed interval.

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