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I just have a quick question about bump functions. If we're dealing with $C^{\infty}_{0}((0,\infty))$, i.e. all smooth bump functions on $(0,\infty)$, obviously any $f\in C^{\infty}_{0}((0,\infty))$ vanishes at infinity, but does it also have to vanish at 0 in order to have compact support?

Thank you!

TinaBelcher
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  • Perhaps the better thing to ask yourself is what compact sets look like on $(0,\infty)$. Are they the same as on $\Bbb R$ or do they differ somehow because of the relative topology? (Hint: both notions of compactness are the same so $(0,a]$ cannot be compact which means that compact sets have to stay away from $0$.) – Cameron Williams Oct 24 '14 at 04:54
  • Great, thank you for the response. That's pretty much what I thought. I was getting a little confused because on wikipedia it says that the support of $f$ is the closure of the subset of the domain where $f$ is nonzero. In this case, I couldn't take the closure of $(0,a]$, so I assumed a support of $(0,a]$ couldn't be compact but wasn't 100% sure. – TinaBelcher Oct 24 '14 at 05:10

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