1. ## Subdividing an interval

Suppose I have a closed, bounded, non-degenerate interval $\displaystyle I = [a, b]$.
I have a set X initially containing the points a and b.

My first action is to find the midpoint of I, call it c, and add c to X.

I now have two new intervals $\displaystyle I_{1}=[a, c]$ and $\displaystyle I_{2}=[c, b]$.

My next action is to find the midpoints of $\displaystyle I_{1}$ and $\displaystyle I_{2}$, lets call them $\displaystyle c_{1}, c_{2}$ and add them to X.

I now have four intervals.... and I continue this procedure until I can subdivide no more.

Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?

2. ## Re: Subdividing an interval

Originally Posted by director
Suppose I have a closed, bounded, non-degenerate interval $\displaystyle I = [a, b]$.
I have a set X initially containing the points a and b.
My first action is to find the midpoint of I, call it c, and add c to X.
I now have two new intervals $\displaystyle I_{1}=[a, c]$ and $\displaystyle I_{2}=[c, b]$.
My next action is to find the midpoints of $\displaystyle I_{1}$ and $\displaystyle I_{2}$, lets call them $\displaystyle c_{1}, c_{2}$ and add them to X.
I now have four intervals.... and I continue this procedure until I can subdivide no more.
Does X contain all points in the original interval I? Or stated differently, is there a point in I that is not in X?
What makes you think that "I can subdivide no more?" At each stage your set X is only finite.

Have you studied the famous Cantor Set ?

3. ## Re: Subdividing an interval

Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies $\displaystyle a\le p\le b$. For any point c, either $\displaystyle x\le c$ or $\displaystyle c\le x$ or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.

4. ## Re: Subdividing an interval

Originally Posted by HallsofIvy
Ignoring the point about "subdivide no more", isn't it obvious that the set of all subintervals does include every point in the original interval? Ever point, p, in [a, b] satisfies $\displaystyle a\le p\le b$. For any point c, either $\displaystyle x\le c$ or $\displaystyle c\le x$ or both. So if x is in [a, b], either x is in [a, c] or x is in [c,b]. In either case, x is in [a, c]U[c, b]. The general proof can be done by induction on the number of halvings.
Originally Posted by director
Suppose I have a closed, bounded, non-degenerate interval $\displaystyle I = [a, b]$.
I have a set X initially containing the points a and b.

My first action is to find the midpoint of I, call it c, and add c to X.

I now have two new intervals $\displaystyle I_{1}=[a, c]$ and $\displaystyle I_{2}=[c, b]$.

My next action is to find the midpoints of $\displaystyle I_{1}$ and $\displaystyle I_{2}$, lets call them $\displaystyle c_{1}, c_{2}$ and add them to X.
At each stage the interval is not added to the set only the new midpoints.

Then we form a partition of $[a,b]$ with that new set of points and find the midpoints of each new subinterval.

So after each stage we have a set of a finite collection.