closure

**mathmad** · November 24th, 2009, 01:00 AM

how

**Shanks** · November 24th, 2009, 01:29 AM

The conclusion is not true if f is not continous function.
If the contious condition is assumed, then it follows immediately from the continuousity of continous function.

**mathmad** · November 24th, 2009, 02:22 AM

I am sorry

The function f is continuous F:X tends to Y
and X and Y are two complete metric spaces

Thank you

**Focus** · November 24th, 2009, 02:31 AM

Suppose that V is a subset. By the continuity of f we have $f^{-1}(\overline{f(V)})$ is closed in X (f^-1 from Y to X maps closed/open sets to closed/open sets resp.).
Notice that $f(V) \subset \overline{f(V)} \implies V\subset f^{-1}(\overline{f(V)})$ . As the closure is the smallest closed superset of V we have $\overline{V}\subset f^{-1}(\overline{f(V)})$ . Now take f of both sides to get the result.

You don't need completeness (or the fact that it is a metric space).

**mathmad** · November 24th, 2009, 02:46 AM

........

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