11.4 Continuity and Connectedness 161Example 11.3.4 f(x) = -,E = (0,1) C R. Let us show that f is not uni-
formly continuous on E but continuous on E: given e > 0, let S > 0 be chosen.
Let x € E and \x — xo\ < 6.
If x 0 — 6 > 0, then \x — x 0 \ < 5 <£> x 0 - 8 < x < x 0 + 5.1 1
X XQ<jx 0 -x|<J^<^5 <£^s< exl
XXQ XXQ (XQ — 5)XQ 1 + EXQHence, f is continuous at XQ and 6 depends on e and Xo. However, dependence
on XQ does not imply that f is not uniformly continuous, because some other
calculation may yield another 6 which is independent of XQ. So, we must show
that the negation of uniform continuity to hold:3e > 0 3 V<5 > 0 3xi,x 2 e E 3 |xj - x 2 \ < S but |/(xi) - f(x 2 )\ > e.Let e = 1. Let S be given. If 5 <\, one can find k 3 5 < -A-^ i.e. k = [| — 1].
Thus, k>2.Letxi=5, x 2 = 6 + £ => 0 < xi < |, 0 < x 2 < 25 < § < 1 =>
Xl,x 2 E E. \xi - x 2 \ = f < f < 6, I/On) - f{x 2 )\
i-rii(S/k)
W+T)
Sik+D > 1. //(5 > | => Let 5' = |. Find x\,x 2 3 \x% — x 2 \ < 5' < 5 and
Ifix^-fix^lKe.
Theorem 11.3.5 Let (X,dx) be a compact metric space, (Y, dy) be a metric
space, and f : X H-> Y be continuous on X. Then, f is uniformly continuous.Remark 11.3.6 Let 0 ^ E C R be non-compact. Then,
(a) 3 a continuous f : E —> R which is not bounded. If E is noncompact then
either E is not closed or not bounded. If E is bounded and not closed,
then E has a limit point x$ 3 XQ £ E. Let f(x) = J- , Vx G E. If E is
unbounded then let f{x) = x, Vx G E.
(b) 3 a continuous bounded function f : E —» R which has no maximum.
If E is bounded let x§ be as in (a). Then, f(x) — 1+/x^1 _x ^, Vx G E.
sup f{x)=l but 3 no x G E 3 f{x) = 1.
(c) If E is bounded, 3 a continuous function f : E —• R which is not
uniformly continuous. Let XQ be as in (a). Let f{x) = —^—, Vx G E
which is not uniformly continuous.11.4 Continuity and Connectedness
Theorem 11.4.1 Let (X,dx),(Y,dy) be metric spaces, 0 ^ E G X be con-
nected and let f : X >-> y be continuous on X. Then, f(E) is connected.