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The question is also posted here.

Let $M=\mathbb{R}$ and $\tau_M=\lbrace U\cup A: U$ open in $\mathbb{R}, A\subset \mathbb{R} \setminus B\rbrace$, where $B$ is a Bernstein set. Then $(M,\tau_M)$ is a topological space called the Michael Line. It is a regular Lindelof space.

Submetrizable = if we can choose a coarser topology on the space $X$ and thus make it a metrizable space.

Let $f: M \to X$ be any one-to-one and onto continuous mapping. Then is $X$ always submetrizable?

Thanks for your help.

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Why must we consider the homeomorphic image rather than the $M$ itself? Is there any background? – Henry.L May 4 '13 at 3:58
Buzyakova posted a new definition of absolutely submetrizable (= every Tychonoff subtopology is submetrizable) in the paper: On absolutely submetrizable spaces – Paul May 4 '13 at 8:43

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