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  1. Could you tell me how part (a) follows from the inequalities (first red box)?
  2. Could you tell me why it is not $f(f^{-1}(E)) = E$ (second red box)?

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1 Answer 1

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  1. Suppose $\mathbf{f}$ is continuous and we want to show that $f_j$ is continuous. Fix $\epsilon>0$, then by continuity of $\mathbf{f}$ there exists $\delta$ such that $|x-y|<\delta$ implies $|\mathbf{f}(x)-\mathbf{f}(y)|<\epsilon$. By chain of inequalities, $$ |f_j(x)-f_j(y)|\leq|\mathbf{f}(x)-\mathbf{f}(y)|<\epsilon. $$ Thus $f_j$ is continuous for each $j=1,\dots,k$. Can you write out the converse?

  2. What happens if $f$ is not surjective?

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