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Question Number 79081 by arkanmath7@gmail.com last updated on 22/Jan/20

I dont know how to show the functions of   this type if it is continuous or not in topological space.  I know the method is to use one of two properties:  the inverse image of eah open is open,  and the inverse image of each closed is closed  but i dont know how to use them.  i will be thaked if someone help ed me with the steps    if f:(R,U) −> (R,U) defined by    f(x) =  { (((1/3) x +1   , x>3)),(((1/2) (x+5)  , x≦3)) :}  Is f continuous?

Idontknowhowtoshowthefunctionsofthistypeifitiscontinuousornotintopologicalspace.Iknowthemethodistouseoneoftwoproperties:theinverseimageofeahopenisopen,andtheinverseimageofeachclosedisclosedbutidontknowhowtousethem.iwillbethakedifsomeonehelpedmewiththestepsiff:(R,U)>(R,U)definedbyf(x)={13x+1,x>312(x+5),x3Isfcontinuous?

Answered by mind is power last updated on 22/Jan/20

Since IR is metrical we can use sequencese  metric space ⇒ separabal ⇒unicite of limites  U_n =3−(1/n)→3,V_n =3+(1/n)→3, withe n∈N^∗   U_n ≤3,V_n >3  f(U_n )=(1/2)(3−(1/n)+5)=4−(1/(2n))→4  f(V_n )=(1/3)(3+(1/n))+1=2+(1/(3n))→4  since lim_(n→∞) f(U_n )#lim_(n→∞) f(V_n )⇒f is note vontinus over 3

SinceIRismetricalwecanusesequencesemetricspaceseparabaluniciteoflimitesUn=31n3,Vn=3+1n3,withenNUn3,Vn>3f(Un)=12(31n+5)=412n4f(Vn)=13(3+1n)+1=2+13n4You can't use 'macro parameter character #' in math mode

Commented by arkanmath7@gmail.com last updated on 22/Jan/20

Commented by arkanmath7@gmail.com last updated on 22/Jan/20

I wrote is it cont? but the qustn in test says show it is cont.

Iwroteisitcont?butthequstnintestsaysshowitiscont.

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