ak416
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Ok i know it should be because it has the same cardinality as R and R is uncountable. But take for example (0,1). Heres a method I would use to count all its elements:
0.1,...,0.9
0.01,...,0.99
0.001,...,0.999
.
.
.
0.(n-1 zeros)1,...,0.(n 9's)
.
.
.
so count starting from the top left and keep going right till you reach the end of the line then count the in the same direction for the next line and so on.
Doesnt this define a bijection from (0,1) to Z+ ? Take any number in (0,1). If it has n digits after its decimal, go down n lines to find its corresponding integer (it will be 9+99+...+(n-1 9's) + the number * 10^n )
Whats wrong with my reasoning?
0.1,...,0.9
0.01,...,0.99
0.001,...,0.999
.
.
.
0.(n-1 zeros)1,...,0.(n 9's)
.
.
.
so count starting from the top left and keep going right till you reach the end of the line then count the in the same direction for the next line and so on.
Doesnt this define a bijection from (0,1) to Z+ ? Take any number in (0,1). If it has n digits after its decimal, go down n lines to find its corresponding integer (it will be 9+99+...+(n-1 9's) + the number * 10^n )
Whats wrong with my reasoning?