Question

Let f:R → R be the identity function f (x)=x. Then f is:
a) continuous but not open.
b) open but not continuous
c) homeomorphism
d) neither open nor continuous​

Answer 1

Step-by-step explanation:

Let f(xni) be a convergent subsequence of f(xn) with limit p. For all i let yi=xni if f(xni)≠p. Otherwise, let r=d(xni,x). Then Br(xni) is open, so f(Br(xni)) is open as well. So f(Br(xni))∩Br(p)∖{p} is not empty. So we can choose yi∈Br(xni) such that f(yi)∈Br(p)∖{p}.

We find yi→x and f(yi)→p and f(yi)≠p for all i. Let S={yi:i∈N}. We find that p is a limit point of f(S) not contained in f(S). However, C=S∪{x} is closed, so f(C) is also closed, and hence contains p. We conclude f(x)=p, so f(xni)→f(x).

Assume for the contrary that f(xn)↛f(x). Then there is a subsequence of f(xn) that always stays a certain distance from f(x). By the Bolzano Weierstrass theorem, this subsequence itself has a convergent subsequence. By the previous observation, this subsequence converges to f(x). This contradicts the fact that it always stays a certain distance from f(x). □

If you manage to prove that such a function does not exist, it might be neat to also look at how general the domain and codomain of f can be made. For example, all arguments in the proposition still work for f:X→Y with any metric space X, and any finite dimensional vector space Y. However, with tweaking the arguments only a bit you find the following.

Let X be a first countable Hausdorff topological space, and let Y be a first countable topological space with no isolated points. Let f:X→Y be open and closed. If xn→x and f(xn) is contained in some sequentially compact set, then f(xn)→f(x).

Answer 2

Step-by-step explanation:

neither open nor continuous