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No. Projection from $\mathbb R^2$ onto $\mathbb R$ is open, but the image of a $G_\delta$ set can be any analytic set. So there are non-Borels among them. See http://en.wikipedia.org/wiki/Analytic_set edit Not a counterexample from a space to itself as required. So a counterexample has to be an open map $\mathbb R^n$ to itself, but not at-most-countable-to-one, since those maps do preserve Borel, as I recall. |
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No. Projection from $\mathbb R^2$ onto $\mathbb R$ is open, but the image of a $G_\delta$ set can be any analytic set. So there are non-Borels among them. See http://en.wikipedia.org/wiki/Analytic_set |
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