Suppose $f:U\subset X_1\to X_2$ is a continuous map from an open subset of $X_1$ into a banach space $X_2$, $f$ is not necessarily onto. Is it true that for any open set $O\subset X_2$, $f^{-1}(O)$ is open?
Continuous map on Banach space
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real-analysis
functional-analysis
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1The usual definition of $f: U \rightarrow X_2$ being continuous is that for $O \subseteq X_2$ open, $f^{-1}(O)$ is open in $U$ (hence in $X_1$). – 2017-02-21
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0is it the same as for any sequence $x_n\to x\in U$ we must have $f(x_n)\to f(x)$? – 2017-02-21
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1Yes. A function $g:A \rightarrow B$ between metric spaces if continuous if and only if the preimage of open sets is open, if and only if whenever $x_n \to x, g(x_n) \to g(x)$ – 2017-02-21