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Let $\Omega$ be an open subset of $\mathbb{R}^n$. Can $BV(\Omega)$ be first countable? If yes, do I have to assume something more for $\Omega$ than just being open?

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I might be missing something, but isn't your question answered by the material at en.wikipedia.org/wiki/Bounded_variation and en.wikipedia.org/wiki/First-countable_space ? –  Yemon Choi Sep 13 '11 at 22:33
    
Hint: Banach spaces. –  David Roberts Sep 14 '11 at 0:00
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