Please show work Thank you Show that a 0 infinity R b a b

Please show work. Thank you.

Show that: (a) |(0. infinity)| = |R| (b) |[a, b]] = |[0.1]| for any a, b with a

Solution

a> |(0,infinity)| = |R|

here we need to show that a function in the open interval (0 , infinity) and R have the same cardinality.

Here we\'ll try to develop a bijective function by mapping the real numbers |R| to (0, infinity).

Let us consider the function f(x) = e^x in order to prove our point. The domain of e^x is all real numbers that is x E R.

And the range of this function is f(X) E (0 , infinity ).

Let e^p = e^q.

=> ln(e^p) = ln(e^q).

or we can say p*ln(e) = q*ln(e).

That is p = q.

Hence, f(x) is injective.

Let r > 0.

=> since e^(ln(r)) = r.

and , r > 0, ln(r) is defined.

=> f(ln(r)) = r.

Therefore, f(x) is surjective as well. Then f is bijective.

So we can say that (0, infinity) an R have the same cardinality.

or |(0,infinity)| = |R|

b> |[a,b]| = |[0,1]|

here we need t find a homomorphism

g : [a,b] --> [0,1]

let say a < x < b and 0 < g(x) < 1

and the map g(x): [a,b][0,1] be

y= g(x) = (x-a)/(b-a)

the above mapping is continuous an one to one

and the inverse of g(x) is

g^(-1)(x) = x(b-a) + a and this proves that g(x) is homomorphic

=> [a,b] is homomorphic to [0,1]

or

|[a,b]| = |[0,1]|

c>

|R| = |[0,1]|

lets assume that we have two maps p:AB and q:BC, both of which show homeomorphisms.

so we can say that clear that pq : AC is again a homeomorphism.

Now we can use this to prove our point,

lets choose a finite open interval (a,b) , and try prove (a,b) is homeomorphic to R

.Then take another open interval (c,d), and show a homeomorphism between (c,d) and (a,b)

, If we take up the interval (0,1) and its image under the function

lets say y = tan(pi*(x12)).

This is clearly a homeomorphism.

Now we\'ll homeomorphically map the open intervall to (0,1) .

hence we can say that |R| = |[0,1]|