Let $f:X\to Y$ be continuous. Show that if $X$ has a countable dense subset, then $f(X)$ satisfies the same condition.
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See also: Continuous image of separable space. – Martin Sleziak Nov 07 '19 at 09:04
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HINT: This is a case in which the most obvious thing to try does work. $X$ is separable, so it has a countable dense subset $D$. Show that $f[D]$ is a countable dense subset of $f[X]$. This is extremely easy if you use the fact that $f$ is continuous if and only if $f^{-1}[U]$ is open in $X$ for each open $U\subseteq Y$.
Brian M. Scott
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Hehe, I like the "This is a case in which the most obvious thing to try does work." ^^ (+1) – C-star-W-star Nov 28 '14 at 22:36