I want to know if the following is true. Let $X$ and $Y$ be topological spaces and $f\colon X\to Y$ a continuous open surjection. Suppose that $X$ is meager, then $Y$ is meager.

Recall that a meager set is a countable union of nowhere dense sets, and a set is nowhere dense if the interior of its closure is empty.

Any help will be appreciated.

  • $\begingroup$ What is the definition of 'meager set'? $\endgroup$
    – Babai
    Sep 28 '15 at 6:07
  • 1
    $\begingroup$ @Babai I just wrote the definitions in the body of the question. $\endgroup$ Sep 28 '15 at 9:37
  • $\begingroup$ So if you say that $X$ is meager, you mean that $X$ is meager in $X$ (i.e., the space $X$ is a countable union of nowhere dense subsets of $X$)? $\endgroup$ Sep 28 '15 at 11:08

I don't think it is true,

Let $X= \mathbb{Q}$ which is clearly meager & $Y$ is singleton. and let $f:X\to Y$ be constant map, so has satisfied all the desired properties, but $Y$ is not meager, because any set is is open in $Y$ since it is singleton.


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