Show that the subspace a b of r is homeomorphic with 0 1 and

Show that the subspace (a, b) of r is homeomorphic with (0, 1) and the subspace [a, b] of R is homeomorphic with [0, 1].

Solution

Show that the subspace (a,b) of R is homeomorphic with (0,1) and the subspace [a,b] of R is homeomorphic with [0,1].

METHOD 1---

Think of mapping the points [not the intervals] (a,0) to (b, 1);

we are mapping endpoints to endpoints of the intervals.

Slope m = (1 - 0)/(b - a) = 1/(b - a).
Hence, the desired line is y = (1/(b - a)) (x - a) = (x - a)/(b - a).

In summary, letting f : (a, b) --> (0, 1) by f(x) = (x - a)/(b - a),
we have f(a) = 0 and f(b) = 1.
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Now, we show that f is a homeomorphism.

(i) Clearly f is continuous (being a polynomial).

(ii) f is onto.

Given y in (0, 1), we want to find x in (a, b) such that f(x) = b.

[Scratch work to find \'a\': (x - a)/(b - a) = y ==> x = (b - a)y + a.]

Let x = (b - a)y + a.
Since y is in (0, 1), and b - a > 0, we have x is in ((b-a) 0 + a, (b-a) 1 + a) = (a, b),
and f(x) = y, as required.

(iii) f is 1-1.
Suppose there exist x, x\' in (a, b) such that f(x) = f(x\').
==> (x - a)/(b - a) = (x\' - a)/(b - a)
==> x - a = x\' - a
==> x = x\'.

(iv) f¹ : (0, 1) --> (a, b) is also continuous.

As a side result of our onto calculations, f¹(y) = (b - a)y + a.
Since this is also a polynomial, f¹ is also continuous.

[If necessary, it is easy to check that (f o f¹)(y) = y, and (f¹ o f)(x) = x.]

METHOD 2---

We need to find a homeomorphism f : (a,b) ->(0,1) and g : [a,b] -> [0,1].

Let a < x < b and 0 < y = f (x) < 1 and the map f : (a,b) -> (0,1) be

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

This map is one-to-one, continuous, and has inverse f ^-1(y) = a + (b-a)y = x and hence a homeomorphism.

therefore,(a,b) is homeomorphic to (0,1).

Use the same map for g as a homeomorphism from [a,b] to [0,1]