Answer to Question #170517 in Real Analysis for Prathibha Rose

Question #170517

Let f:R2 to R defined by f(x)=√(x2+y2) show that f is continuous on R2

1
Expert's answer
2021-03-23T14:34:16-0400

f(x,y)=(x2+y2)f(x,y)=\sqrt{(x^2+y^2)}


now, to show that f(x,y) is continuous everywhere on R2R^2 ,

we first fix an arbitrary (a.b)(a.b) and ϵ>0ϵ>0 .


here note that ff is indeeded the Euclidean's norm on R2R^2 ,

so we can use the reverse triangle inequality.

For any (x,y)R2(x,y)∈R^2 with (x,y)(a,b)<δ=ϵ|(x,y)−(a,b)|<δ=ϵ


d((fx,y),f(a,b))=f(x,y)f(a,b)d((fx,y),f(a,b))=|f(x,y)−f(a,b)|

=x2+y2a2+b2=f(x,y)f(a,b)(x,y)(a,b)<ϵ=|\sqrt{x^2+y^2}-\sqrt{a^2+b^2}|\\ =|f(x,y)|−|f(a,b)||\\ ≤|(x,y)−(a,b)|\\ <ϵ


hence, proving ff is continuous in our arbitrary chosen (a,b).

since, this is our arbitrary chosen ,

we conclude that ff is continuous everywhere.

this assumes that both domain and co-domain of ff

are equipped with corresponding euclidean metrics.


hence,proved.


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