# $\emptyset$ and Topology

Could someone please clarify the following to me?

Suppose $f:X\to Y$ is a surjection from a topological space, $X$, to some set, $Y$. Then is it necessarily true that $f^{-1}(\emptyset)=\emptyset$?

I think this is true, but I want to be sure. Thank you.

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You've defined a function on $X$ and are essentially asking 'what points in $X$ don't map to anything?'. $f$ wouldn't be a function on $X$ if it didn't map each element of $X$ to something – Jonathan May 5 '12 at 11:24
@Jonathan: Thank you very much! – Bumbry May 5 '12 at 11:32
I'd also point out that there is a big difference between $f^{-1}(\emptyset)$ and $f^{-1}(\{ \emptyset \} )$, in case this causes you confusion in the future (i.e the latter makes sense if $\emptyset$ is a member of $Y$) – Jonathan May 5 '12 at 15:02
@Jonathan: Thanks, would you mind saying a little bit more? I am new to this and I don't think I quite know the difference. Aren't they the empty set? What do you mean by member? – Bumbry May 5 '12 at 15:09
If you have a set, the "things" it contains are called its members. You can have a set like $X = \{1 ,2 ,3 \}$, or $Y = \mathbb N$, or anything really, like $Z = \{\mathrm{yellow, \ blue, \ green, \ banana}, \ 5 \}$; $1$ is a member of $X$, $28493$ is a member of $Y$, $\mathrm{blue}$ is a member of $Z$. It's possible for the empty set to be a member of a set e.g. $X = \{ \mathbb N , \mathbb Z , \mathbb R , \emptyset \}$. This is a set whose members are themselves sets. – Jonathan May 5 '12 at 16:23