Continuous functions

1. Continuous functions Definition: Let f be a function from a metric space (X,d1) to a metric space (Y,d2) then we say that f is a continuous function at x=a if : for each open set G in Y such that f(a)G, then f-1(G) is open set in X. f is continuous if it continuous at all aX. Example: Let f:R→R be a function defined by: f(x)= Show that f is not continuous at x=0 but it continuous at x>0. Proof: The interval V= is open set And f(0)=0V But: f-1(V)={xX:f(x)V}={xR:f(x) }

2. f-1(V)={0} since {0} is not open set in R thus f-1(V)={0} is not open set in R therefore f is not continuous at x=0 Now for all x>0 Let V be an open set such that xV f-1(V)= Hence f-1(V) is open set Therefore f is continuous function Theorem: A function f:(X,d1)(Y,d2) is continuous iff f-1(F) is closed set in X for each closed set F in Y.

3. Proof: () suppose that f is continuous function Let F be a closed set in Y Thus Fc isopen set in Y Hence f-1(Fc) is open set in X But f-1(Fc)=f-1(X-F)= X-f-1(F) Therefore f-1(F) is closed in X () Suppose that f-1(F) is closed set in X for each closed set F in Y. Let A be an open set in Y, then Ac is closed set in Y By hypothesis we have f-1(Ac) is closed in X But f-1(Ac)=f-1(X-A)= X-f-1(A) Hence A is open set in X Therefore f is continuous function.

4. Theorem: Let f: (X,d1)(Y,d2) be a function and  AX , then f is continuous function. Proof: To prove that f is continuous by the theorem { A function f:(X,d1)(Y,d2) is continuous iff f-1(F) is closed set in X for each closed set F in Y}. Let F be a closed set in Y. By the properties of the closure of the set F we have To prove that f-1(F) is closed set in X we need prove that: {by theorem A closed set iff Let  x (because f is a function) Now {because } {because f(f-1(F))F}

5. Thus f(x)F x f-1(F) Hence Therefore f-1(F) is closed set in X Thus F is continuous function. Theorem: Let f: (X,d1)(Y,d2) be a continuous function and (X,d1) be a compact space then f(X) is compact set, such that f(X)={f(x):xX} is a range of f. Proof: Let {G:}be an open cover of f(X).

6. Since f is a continuous function then f-1(G) is open set in X. Thus {f-1(G):} be an open cover of X. Since X is compact set then this cover has finite open cover such {f-1( ): i=1,2,…,n} . Hence { : i=1,2,…,n} is finite cover of f(X). Therefore f(X) is compact set in Y.

7. Corollary: Let f: (X,d1)(R,d2) be a continuous function and (X,d1) be a compact space,(R,d2) be the usual space then f is bounded (Hence f(X) is bounded set) Proof: Since f is continuous and X is compact then: f(X) is compact set in (R,d2) by theorem { Let f: (X,d1)(Y,d2) be a continuous function and (X,d1) be a compact space then f(X) is compact} Thus by theorem (Heine-Borel) we have f(X) is closed and bounded. Hence f(X) is bounded Therefore f is bounded function. Theorem: Let f:X→R be a continuous function and X is compact set then there exists x0,y0X such that f(x0) ≤ f(x) ≤ f(y0) ,  xX. Hence : Every continuous function with compact domain X , it has maximal and minimal points. Theorem(Intermediate value property) : Let f: [a,b]R be a continuous function then: For all x1 ,x2 [a,b] and for all s between f(x1),f(x2) ,there exists z between x1 , x2 such that f(z)=s.

8. Corollary-1: Let f: [a,b]R be a continuous function such that f(-x)=-f(x){f is odd function} Then there exists t[a,b] such that f(t)=0. Proof: We have two cases: 1-If f(x)=0 for all x[a,b] then the proof is complete. 2- if f(x)0 then there exists y[a,b] such that f(y) 0 Now a- If f(y)>0 then f(-y)=-f(y)<0 thus f(-y) < 0 < f(y) by theorem (Intermediate value property) there exists t[a,b] such that f(t)=0. B- If f(y)<0 then f(-y)=-f(y)>0 thus f(y) < 0 < f(-y) by theorem (Intermediate value property) there exists t[a,b] such that f(t)=0.

9. Corollary-2: Let f: [a,b][a,b] continuous function then there exists t[a,b] such that f(t)=t. Proof: Suppose that : g: [a,b]R is a function defined by: g(x)=f(x)-x since f is continuous function then g is continuous function also. We note that f(a),f(b)[a,b] and a  f(a),f(b)  b Thus a  f(a)  0  f(a)-a= b  f(b)  0  f(b)-b= g(a) g(b) Hence  0  By theorem (Intermediate value property) there exists t[a,b] such that: g(t)=0 f(t)-t=0  f(t)=t.