1.5.CONTINUOUS FUNCTIONS AND DISTANCE IN METRIC SPACE 11 ce of nested balls B( )C B( 1).mor equences and it cor rerges to. Using the Baire category theorem,we get another proof that 0,I is uncountable. Gs set,countable union of closed sets is called an 0ion7.Phere dow not erist a mction加.广：R→hich t continous ony t7 We need a lemma first Lemma 1.8.The points of continuity of f is a Gs set. Proof.Recall that f is continuous at z if and only if the oscillation wf()=0.Therefore the set of points of continuity of f is ∩wr回)<》 It is easy to show that ()<is open. 0 Q=0G where each G is open set.We write QasQ=),then R=UG:Ufa). =1 neSCeckoad.saupoeteotainsanialaiaroil,thamtdaeethisanopemileralz）cG (z,°GmQ. The only ossible case is =u.Hence Ge is nowhere dens The above expression writes R as a union of countable nowhere dense sets.This contradicts to the Baire category theorem. 1.5 Continuous functions and Distance in metric space Given a function f:ECR"R,f is continuous at rE,if e0,there exists6>0 such that If(y)-fr川≤6，y∈B(x,6)nE. f is called continuous on E if f is continuous at every point of E.This definition does not require E is open. Theorem 1.9.Suppose f:FR be a continuous function defined on a compact set F,then f is uniform continuous and attains its marimum and minimum. 1.5. CONTINUOUS FUNCTIONS AND DISTANCE IN METRIC SPACE 11 D2 c ∩ B(x1, ). Inductively, we get a sequence of nested balls B(xn, n) ⊂ B(xn−1, n−1), moreover we can easily arrange that limn→∞ n = 0. Thus {xn} is a Cauchy sequences and it converges to, say x. Since X = S∞ n=1 Dn, thus x ∈ Dk for some k. However due to the construction x ∈ B(xk, k), which contradicts to that B(xk, k) ∩ Dk = ∅. Using the Baire category theorem, we get another proof that [0, 1] is uncountable. Countable intersection of open sets is called a Gδ set, countable union of closed sets is called an Fσ set. We give a more interesting application of Baire’s category theorem. Proposition 1.7. There does not exist a function f : R → R which is continuous only at all rational numbers. We need a lemma first. Lemma 1.8. The points of continuity of f is a Gδ set. Proof. Recall that f is continuous at x if and only if the oscillation ωf (x) = 0. Therefore the set of points of continuity of f is \∞ n=1 {x|ωf (x) < 1 n }. It is easy to show that {x|ωf (x) < 1 n } is open. Proof of the Proposition. Using the above lemma, it is suffice to show that Q is not a Gδ set. Suppose not, then assume Q = \∞ n=1 Gn, where each Gn is open set. We write Q as Q = {q1, q2, · · · }, then R = [∞ n=1 G c n [∞ i=1 {qi}. Gc n is closed, suppose it contains an interior point, then there exists an open interval (x, y) ⊂ Gc n . Therefore (x, y) c ⊃ Gn ⊃ Q. The only possible case is x = y. Hence Gc n is nowhere dense. The above expression writes R as a union of countable nowhere dense sets. This contradicts to the Baire category theorem. 1.5 Continuous functions and Distance in metric space Given a function f : E ⊂ R n → R, f is continuous at x ∈ E, if ∀ > 0, there exists δ > 0 such that |f(y) − f(x)| ≤ , ∀y ∈ B(x, δ) ∩ E. f is called continuous on E if f is continuous at every point of E. This definition does not require E is open. Theorem 1.9. Suppose f : F → R be a continuous function defined on a compact set F, then f is uniform continuous and attains its maximum and minimum