Show that if f: X->Y is a homeomorphism, then:

$\displaystyle f(\partial(A))=\partial(f(A))$

I am stuck!

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- Feb 24th 2009, 07:02 PMAndreamethomeomorphisms and interior, boundary
Show that if f: X->Y is a homeomorphism, then:

$\displaystyle f(\partial(A))=\partial(f(A))$

I am stuck! - Feb 25th 2009, 01:28 AMaliceinwonderland
I assume A is a subset of X.

Since $\displaystyle \partial A = \overline{A} \cap \overline {X \setminus A}$, $\displaystyle f (\partial A) = f( \overline{A} \cap \overline {X \setminus A})$.

We need to show that $\displaystyle f( \overline{A} \cap \overline {X \setminus A})$ is $\displaystyle \overline{f(A)} \cap \overline {Y \setminus f(A)}$, which is $\displaystyle \partial (f(A))$.

1. For every subset A of X, one has $\displaystyle f(\bar{A}) \subset \overline{f(A)}$ when f is continuous. If f is a homeomorphism, $\displaystyle f(\bar{A}) = \overline{f(A)}$.

2. $\displaystyle f(X \setminus A) = (Y \setminus f(A))$. Using 1, $\displaystyle f(\overline{X \setminus A}) = \overline{Y \setminus f(A)}$.

Now, it remains to combine 1 & 2 to get the answer.