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With Z=f(X), the function f:X→Z is a continuous surjection and we wish to show that Z is separable.
With closure bars denoting closure in X or in Z, we have
f(A¯¯¯¯)⊂f(A)¯¯¯¯¯¯¯¯¯¯⊂Z
because f is continuous. We have
Z=f(X)=f(A¯¯¯¯)
because X=A¯¯¯¯. So we have
Z=f(X)=f(A¯¯¯¯)⊂f(A)¯¯¯¯¯¯¯¯¯¯⊂Z.
Therefore f(A)¯¯¯¯¯¯¯¯¯¯=Z. So f(A) is a countable dense subset of Z.



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