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Let X\, and Y\, be metric spaces, and f:X\to Y\, be a mapping.

(i) Prove that if f^{{-1}}(G)\, is open whenever G\subset Y\, is open, then f\, is continuous.

(ii) Prove that f\, is continuous if and only if f^{{-1}}(F)\, is closed whenever F\subset Y\, is closed.


Suppose f^{{-1}}(G)\, is open whenever G\subset Y\, is open. Let x_{0}\in X\, and \epsilon >0\,. Since the ball B(f(x_{0}),\epsilon )\, is open, so is f^{{-1}}\left[B(f(x_{0}),\epsilon ))\right]\,. Since f(x_{0})\in B(f(x_{0}),\epsilon ),x_{0}\in f^{{-1}}\left[B(f(x_{0}),\epsilon )\right]\,. Since f^{{-1}}\left[B(f(x_{0}),\epsilon )\right] is open there exists a \delta >0\, such that B(x_{0},\delta )\subset f^{{-1}}\left[B(f(x_{0}),\epsilon )\right]\,. This implies f(B(x_{0},\delta ))\subset B(f(x_{0}),\epsilon )\, making f\, continuous at x_{0}\,. Since x_{0}\, was arbitrary, f\, is continuous.


Suppose f^{{-1}}(F)\, is closed whenever F\in Y\, is closed. Let G\subset Y\, be open. Then G^{c}\, is closed. By the hypothesis, f^{{-1}}(G^{c})\, is closed.

If x\in f^{{-1}}(G^{c})\,, then f(x)\in G^{c},f(x)\notin G,x\notin f^{{-1}}(G)\, and x\in \left[f^{{-1}}(G)\right]^{c}\,. Similarly, if x\in \left[f^{{-1}}(G)\right]^{c}\, then x\notin f^{{-1}}(G),f(x)\notin G,f(x)\in G^{c}\,, and x\in f^{{-1}}(G^{c})\,. Therefore f^{{-1}}(G^{c})=\left[f^{{-1}}(G)\right]^{c}\,.

Now, f^{{-1}}(G^{c})=\left[f^{{-1}}(G)\right]^{c}\, is closed, so f^{{-1}}(G)\, is open. So f^{{-1}}(G)\, is open whenever G\subset Y\, is open. From part (i), f\, is continuous.

Suppose f\, is continuous. Let G\subset Y\, be closed. Let \{x_{n}\}\subset f^{{-1}}(G)\, converge to x\in X\,. Since f\, is continuous \{f(x_{n})\}\subset G\, converges to f(x)\in Y\,. Since G\, is closed, f(x)\in G\, and so x\in f^{{-1}}(G)\, which means f^{{-1}}(G)\, is closed (since this is the limit of the arbitrary sequence chosen above). Therefore f^{{-1}}(G)\, is closed whenever G\, is closed.

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