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Not really homework, but a typical exercise question, so I figured it's appropriate to post it here.
X,Y topological spaces
f:X→Y
x is a point in X
Prove that the following two statements are equivalent:
(i) [itex]f^{-1}(E)[/itex] is open for every open E that contains f(x).
(ii) If [itex]\{x_i\}[/itex] is a net such that [itex]x_i\rightarrow x[/itex], then [itex]\{f(x_i)\}[/itex] is a net such that [itex]f(x_i)\rightarrow f(x)[/itex]
None.
(i) implies (ii): Easy. See below.
(ii) implies (i): I haven't been able to prove this. One thing I tried was to let the directed set be the set of open neighborhoods of x, partially ordered by reverse inclusion. [itex](i\leq j\iff j\subset i)[/itex]. I think I ended up proving that [itex]f^{-1}(E)[/itex] contains an open neighborhood of x, but that doesn't prove that [itex]f^{-1}(E)[/itex] is open.
This is how I show that (i) implies (ii):
Let E be an open set that contains f(x), and let [itex]\{x_i\}[/itex] be a net such that [itex]x_i\rightarrow x[/itex]. (i) implies that [itex]f^{-1}(E)[/itex] is open, and we know that it contains x. So there exists an i0 such that
[tex]i\geq i_0\implies x_i\in f^{-1}(E)[/tex]
But the condition on the right is equivalent to [itex]f(x_i)\in E[/itex], so we have [itex]f(x_i)\rightarrow f(x)[/itex].
Homework Statement
X,Y topological spaces
f:X→Y
x is a point in X
Prove that the following two statements are equivalent:
(i) [itex]f^{-1}(E)[/itex] is open for every open E that contains f(x).
(ii) If [itex]\{x_i\}[/itex] is a net such that [itex]x_i\rightarrow x[/itex], then [itex]\{f(x_i)\}[/itex] is a net such that [itex]f(x_i)\rightarrow f(x)[/itex]
Homework Equations
None.
The Attempt at a Solution
(i) implies (ii): Easy. See below.
(ii) implies (i): I haven't been able to prove this. One thing I tried was to let the directed set be the set of open neighborhoods of x, partially ordered by reverse inclusion. [itex](i\leq j\iff j\subset i)[/itex]. I think I ended up proving that [itex]f^{-1}(E)[/itex] contains an open neighborhood of x, but that doesn't prove that [itex]f^{-1}(E)[/itex] is open.
This is how I show that (i) implies (ii):
Let E be an open set that contains f(x), and let [itex]\{x_i\}[/itex] be a net such that [itex]x_i\rightarrow x[/itex]. (i) implies that [itex]f^{-1}(E)[/itex] is open, and we know that it contains x. So there exists an i0 such that
[tex]i\geq i_0\implies x_i\in f^{-1}(E)[/tex]
But the condition on the right is equivalent to [itex]f(x_i)\in E[/itex], so we have [itex]f(x_i)\rightarrow f(x)[/itex].