Let E be a subset o/IRm and f : E —»• IRP a function. Suppose that хб£ and that f is continuous at x. // xn 6 E for all n and xn —»• x as n —> oo, йегг f (xn) —»• f(x) as n —»• oo.
Proof. Left to the reader.
Another way of looking at continuity, which will become progressively more important as we proceed, is given by the following lemma.
Lemma 4.31. The function f : W71 —»• IRP is continuous if and only if f-1^) is open whenever U is an open set in W'.
The reader may need to be reminded of the definition
f-\U) = {xeRm : f(x) eU}.
Proof. As with most of the proofs in this section, this is just a matter of writing things down correctly. We split the proof into two parts.
Necessity Suppose f is continuous and U is an open set in Rp. If x e f_1(6r), then f (x) 6 U. But U is open, so there exists an e > 0 such that 5(f (x), e) С U. Since f is continuous at x, we can find a S > 0 such that
||f(x) — f(y)|| < 6 whenever ||x — y|| < S.
We thus have B(x.,S) С f_1(^)- It follows that f_1(^) is open. Sufficiency Suppose that f_1(6r) is open whenever U is an open subset of W. Let x € Rm and e > 0. Since B(f(x),e) is open, it follows that f_1(B(f(x),e)) is open. But x 6 f_1(B(f(x),e)), so there exists a 8 > 0 such that S(x, 5) С f_1(B(f(x), e)). We thus have
||f(x) — f(y)|| < 6 whenever ||x — y|| < 8.
It follows that / is continuous.