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Continuity preserving Lindelöf property


"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
Good night! Here is the problem:

Let $f: X \to Y$ be continuous. Show that if X is Lindelöf, or if X has a countable dense subset, then $f(X)$ satisfies the same condition.

I don't know how to work with the Lindelöf condition. I've done the countable dense subset, though:

Let $D \subset X$ be the mentioned countable dense subset, therefore we know that $\overline{D} = X$. Using one of the equivalent conditions for a mapping to be continuous, we know that $f( \overline{A}) \subset \overline{f(A)}$ for any $A \subset X$, and combining it with our hypothesis we conclude that $f(\overline{D}) = f(X) \subset \overline{f(D)}$, thus $\overline{f(D)} = f(X)$. The countable assertion follows from the continuity as well.

Just for reference, I know the definition of Lindelöf but I am having trouble proving things with it. Here it is:

Let $X$ be a topological space and $\{ U_{\alpha} \}$ an open covering of $X$. We say that $X$ is Lindelöf when we can find a countable open covering $\{ U_{\alpha_i} \}$ that covers $X$.

Thanks! (Wave)



Active member
Mar 1, 2012
Let [tex]\mathscr{B} [/tex] be an open cover for Y, [tex] \cup \mathscr{B} = Y [/tex]
[tex] f^{-1} \left( \mathscr{B} \right) [/tex] is an open cover for X, we can find a countable subcover of [tex]f^{-1}(\mathscr{B}) [/tex] (since X is lindelof) which covers X the image of this subcover is a countable subcover of [tex]\mathscr{B}[/tex] which covers Y. Y is lindelof