Cardinality of continuous functions

talolard
Messages
119
Reaction score
0

Homework Statement


What is the cardinality of the set of all continuous real valued functions [tex][0,1] \rightarrow R[/tex].




The Attempt at a Solution


In words:
I will be using the Cantor Bernstien theorem. First the above set, let's call it A, is lesser then or equal to the set of all functions from R to R which has a cardinality of [tex]\aleph ^ \aleph[/tex]
Also, the cardinality of all continuous functions [tex][0,1] \rightarrow R[/tex]. is lesser then or equal too the cardinality of the set of all functions, continuous and not continuous, from [tex][0,1] \rightarrow R[/tex].
The cardinality of the segment [0,1] is [tex]\aleph[/tex] and so the cardinality of all functions [tex][0,1] \rightarrow R[/tex]. is also [tex]\aleph ^ \aleph[/tex] and so by the cantor bernsiten theorem the cardinality of A is also [tex]\aleph ^ \aleph[/tex]

In formal math:
I use C to denote continuous functions.
[tex]\aleph ^ \aleph =|R^R| \geq |R^{C[0,1]}| = |C[0,1] \rightarrow R| = |R^{C[0,1]}| \leq |R^{[0,1]}|=\ \aleph ^ \aleph[/tex]


Is this correct? and formal anough?
Thanks
Tal
 
Physics news on Phys.org
Ahhh I see my mistake already.
Is it enough to say that a coninuous function is determined by its values on Q. Therefore
[tex]C[0,1] \rightarrow R = R^{[0,1] \in Q}= \aleph[/tex]
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K