4.8. (Sequences in compact metric spaces have convergent subsequences).

This follows from Theorem 3.38 (b) (see text) in case the range of the sequence is infinite and is obvious if the range of the sequence is finite.

4.13 (If f:[a, b] -> R is continuous and if it is zero over rationals then it is zero over irrationals too)

Let p be an irrational number in [a, b]. Want to show that f(p)=0. Take any sequence of rational numbers converging to p in [a, b] (easy to establish the existence of such a sequence). Since and since f is continuous, it must be that . But all are zero, so that must be zero too.

4.25 (f continuous [a, b] -> R, then between every two maxima, there is a minimum.)

(using the notation in the original statement of the problem; see text). Since f is continuous over [a, b], it is also continuous over ; since the last interval is closed (and compact), by Theorem 4.28 in text, there is a point p in such that (the edges of the interval are excluded since both are local maxima). That point must be a local minimum (in fact, it is where the function has an absolute minimum over ).

4.35 (No continuous 1-1 mapping from [0,1] onto [0,1]x[0,1]).

Otherwise there is a continuous (one-to-one) mapping from [0,1]x[0,1] onto [0,1] (the inverse of the given mapping ). Take any point p in [0,1]x[0,1] that is mapped in the interior of [0,1] (say, to 1/2). Note that since is continuous over [0,1]x[0,1] , it is continuous over any subset of that set. So, is continuous over the set [0,1]x[0,1] without the point p. But this is not possible, since the last set is connected, while the image of it under is [0, 1/2) and (1/2, 1] and it is disconnected (continuous maps send connected sets to connected sets - easy and proven in class).

[Bonus question] (find f:S->T such that f is continuous and 1-1, yet the inverse mapping is not continuous).

Take S=[0, 1) (semiopen interval in R); take T to be a circle in . Let f do the following

(start with [0,1) drawn vertically on the left side, in red; end with T in blue):

This f is continuous and one-to-one, but the inverse mapping is not continuous.