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Here's a problem:

I've done the first 2 parts but I have no idea of how to approach part iii, and I'm very confused for part iv because isn't a singleton set a closed subset?

Any help would be appreciated. Thanks!

Chinnapparaj R
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NaturalLog420
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  • It is not true that any subset of a metric space is open. If you don't care enough to type up the question and show some effort, I don't care enough to click through. – Ross Millikan Oct 18 '17 at 04:23

1 Answers1

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$iii)$ With the function $d_2$ you defined as "metric" on the set of integrable functions on $[0,1]$ you have that both the characteristic function of the 0-singleton, $\chi_{\{0\}}$ and the constant function $0$ are integrable, moreover $d_2(\chi_{\{0\}},0)=0$ but they are not equal. $iv)$Each point in a metric space is closed, finite unions of closed sets are still closed so every set is closed (hence every set is open too)

Tancredi
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