Lemma 0.5.LetX,Ybe metric spaces andf:X→Y. The following areequivalent:(i)fis continuous atx(i.e,(0.9)holds).(ii)f(xn)→f(x)wheneverxn→x.(iii)For every neighborhoodVoff(x),f−1(V)is a neighborhood ofx.Proof.(i)⇒(ii) is obvious. (ii)⇒(iii): If (iii) does not hold, there isa neighborhoodVoff(x) such thatBδ(x)6⊆f−1(V) for everyδ. Hencewe can choose a sequencexn∈B1/n(x) such thatf(xn)6∈f−1(V). Thusxn→xbutf(xn)6→f(x). (iii)⇒(i): ChooseV=Bε(f(x)) and observethat by (iii),Bδ(x)⊆f−1(V) for someδ.The last item implies thatfis continuous if and only if the inverse imageof every open (closed) set is again open (closed).Note: In a topological space, (iii) is used as the definition for continuity.However, in general (ii) and (iii) will no longer be equivalent unless one usesgeneralized sequences, so-called nets, where the index setNis replaced byarbitrary directed sets.Thesupportof a functionf:X→Cnis the closure of all pointsxforwhichf(x) does not vanish; that is,supp(f) ={x∈X|f(x)6= 0}.(0.10)IfXandYare metric spaces, thenX×Ytogether withd((x1,y1),(x2,y2)) =dX(x1,x2) +dY(y1,y2)(0.11)is a metric space. A sequence (xn,yn) converges to (x,y) if and only ifxn→xandyn→y. In particular, the projections onto the first (x,y)7→x,respectively, onto the second (x,y)7→y, coordinate are continuous.In particular, by the inverse triangle inequality (0.1),|d(xn,yn)−d(x,y)|≤d(xn,x) +d(yn,y),