1. If f:X-->Y is a homeomorphism, the f(Cl(A)) = Cl(f(A)) for every $\displaystyle A \in X$

2. If f:X-->Y is a homeomorphism, the f($\displaystyle \mathfrak{d}$(A)) = $\displaystyle \mathfrak{d}(f(A))$ for every $\displaystyle A \in X$

2. Originally Posted by r2dee6
1. If f:X-->Y is a homeomorphism, the f(Cl(A)) = Cl(f(A)) for every $\displaystyle A \in X$

2. If f:X-->Y is a homeomorphism, the f($\displaystyle \mathfrak{d}$(A)) = $\displaystyle \mathfrak{d}(f(A))$ for every $\displaystyle A \in X$
Lemma 1. Let $\displaystyle f:X \rightarrow Y$ be a function on the indicated topological spaces. f is continuous iff for each subset A of X, $\displaystyle f(\overline{A} ) \subset \overline{f(A)}$.

(1) Since $\displaystyle f$ is continuous, $\displaystyle f(\overline{A} ) \subset \overline{f(A)}$by lemma 1. We know that $\displaystyle f^{-1}$ is also continuous. Thus, $\displaystyle f^{-1}(\overline{f({A})}) \subset \overline{f^{-1}f({A})}$ by lemma 1. We have $\displaystyle \overline{f(A)} \subset f(\overline{A})$. We conclude that $\displaystyle f(\overline{A} ) = \overline{f(A)}$.

(2) Since f is a homeomorphism, $\displaystyle f(\mathfrak{d}(A)) = f(\overline{A} \cap \overline{X \setminus A} ) = \overline{f(A)} \cap \overline{Y \setminus f(A)} = \mathfrak{d}(f(A))$.