The Real And Complex Number Systems

from b to A) are disjoint. So, if every line contains a point of T, then it leads us to get T is
uncountable. However, T is countable. So, it has some line (form a to A, and from b to A)
is in S T. So, it means that S T is path-connected. So, S T is connected.
4.45 Given a function f : R n R m which is 1-1 and continuous on R n .IfA is open
and disconnected in R n , prove that fA is open and disconnected in fR n .
Proof: Theexerciseiswrong. There is a counter-example. Let f : R R 2
f
cos 2x
1x
cos 2x
1x
2
2
2x
,1 sin if x 0
1x 2
2x
, 1 sin if x 0
1x 2
Remark: If we restrict n, m 1, the conclusion holds. That is, Let f : R R be
continuous and 1-1. If A is open and disconnected, then so is fA.
Proof: In order to show this, it suffices to show that f maps an open interval I to
another open interval. Since f is continuous on I, andI is connected, fI is connected. It
implies that fI is an interval. Trivially, there is no point x in I such that fx equals the
endpoints of fI. Hence, we know that fI is an open interval.
Supplement: Here are two exercises on Homeomorphism to make the reader get more
and feel something.
1. Let f : E R R. Ifx, fx : x E is compact, then f is uniformly continuous
on E.
Proof: Let x, fx : x E S, and thus define gx Ix x, fx : E S.
Claim that g is continuous on E. Consider h : S E by hx, fx x. Trivially, h is 1-1,
continuous on a compact set S. So, its inverse function g is 1-1 and continuous on a
compact set E. The claim has proved.
Since g is continuous on E, we know that f is continuous on a compact set E. Hence, f
is uniformly continuous on E.
Note: The question in Supplement 1, there has another proof by the method of
contradiction, and use the property of compactness. We omit it.
2. Let f : 0, 1 R. Ifx, fx : x 0, 1 is path-connected, then f is continuous
on 0, 1.
Proof: Let a 0, 1, then there is a compact interval a a 1 , a 2 0, 1. Claim
that the set
x, fx : x a 1 , a 2 : S is compact.
Since S is path-connected, there is a continuous function g : 0, 1 S such that
g0 a 1 , fa 1 and g1 a 2 , fa 2 . If we can show g0, 1 S, wehaveshown
that S is compact. Consider h : S R by hx, fx x; h is clearly continuous on S. So,
the composite function h g : 0, 1 R is also continuous. Note that h g0 a 1 ,and
h g1 a 2 , and the range of h g is connected. So, a 1 , a 2 hg0, 1. Hence,
g0, 1 S. We have proved the claim and by Supplement 1, we know that f is
continuous at a. Sincea is arbitrary, we know that f is continuous on 0, 1.
Note: The question in Supplement 2, there has another proof directly by definition of
continuity. We omit the proof.
4.46 Let A x, y :0 x 1, y sin 1/x, B x, y : y 0, 1 x 0,
and let S A B. Prove that S is connected but not arcwise conneceted. (See Fig. 4.5,