How would you prove this statement? (0,1) and [0,1) have the same cardinality. H

Question

How would you prove this statement? (0,1) and [0,1) have the same cardinality.
How would you prove this statement? (0,1) and [0,1) have the same cardinality.

Explanation / Answer

(0,1) --> [0,1) is trivial: Just map x --> x; [0,1) --> (0,1): We know that the half-open interval [1/2,1) is a subset of (0,1). So if we can map [0,1) to [1/2, 1) so that each different value from the [0,1) maps to a SEPARATE value in [1/2, 1), we are done. But this is easy: We "contract" [0,1) to [0,1/2) by computing y/2; and then "move it to the right" by adding 1/2. So, the mappping is: y --> 1/2 + y/2. Formally, we now have to show that two different y's map to two different values; which is logically the same as "two equal mapping result originate from two equal y's" Pf. 1/2 + y/2 = 1/2 + z/2; y/2 = z/2; y = z; The mapping result is actually a subset of (0,1): So, for y ? [0,1), 1/2 + y/2 ? (0,1): y ? [0,1) is the same as 0 = y < 1, from where we continue as follows: 0 = y < 1; 0 = y/2 < 1/2; 1/2 = 1/2+y/2 < 1; 0 < 1/2 = 1/2+y/2 < 1; 0 < 1/2+y/2 < 1; The last line is the same as 1/2 + y/2 ? (0,1), so we are done and the proof is closd.

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