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Define an equivalence relation on $\mathbb{R}$ (with the standard topology) such that the quotient space is 3 points with the indiscrete (trivial) topology.

KangHoon You
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HINT: Let $X=\{0,1,2\}$ have the indiscrete topology. Find a map $f:\Bbb R\to X$ such that if $A\subseteq X$, then $f^{-1}[A]$ is open in $\Bbb R$ if and only if $A=X$ or $A=\varnothing$. Note that this will be the case if $f^{-1}[\{x\}]$ is neither closed nor open for each $x\in X$. (Why?)

Brian M. Scott
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  • Shouldn't I be looking for an equivalence relation, not a function? – KangHoon You Oct 15 '13 at 04:40
  • @KangHoonYou: Yes; use the function to give you the equivalence relation. If you identify the points $0,1$, and $2$ with the three points of your quotient space, the map $f$ is essentially the quotient map. – Brian M. Scott Oct 15 '13 at 04:41
  • @KangHoon see http://math.stackexchange.com/a/526393/622 – Asaf Karagila Oct 15 '13 at 04:46
  • @BrianM.Scott does the following map work. If not, can you explain why...f(x)={x+3 if x=0,1,2,; x if x$\neq$0,1,2 – KangHoon You Oct 16 '13 at 02:20
  • @KangHoonYou: That isn’t a function from $\Bbb R$ to $X$. If $x=5$, for instance, you have $f(x)=5$, which isn’t in $X$. – Brian M. Scott Oct 16 '13 at 02:25
  • @BrianM.Scott Oh okay! How about this? f(x)={0 if x=0, 1 if x<0 and 2 if x>0} – KangHoon You Oct 16 '13 at 02:33
  • @KangHoonYou: It’s a function from $\Bbb R\to X$, but the partition of $\Bbb R$ that it induces won’t give you the indiscrete topology on the quotient space. You really do need to make sure that $f^{-1}[{x}]$ is neither open nor closed for each $x\in X$. – Brian M. Scott Oct 16 '13 at 02:35
  • @BrianM.Scott Should I use a projection function? I'm really not sure where to go with this. – KangHoon You Oct 16 '13 at 02:47
  • @KangHoonYou: I can’t say much more than I’ve already said without simply working the problem. Your equivalence relation $\sim$ on $\Bbb R$ is of course going to be that $\alpha\sim\beta$ iff $f(\alpha)=f(\beta)$, and you just have to choose $f$ to give the desired result. $f$ will in essence be the quotient map and $X$ the quotient space. At this point I’d suggest that concentrate on answering the Why? in my answer: if you can’t answer it, the problem most likely is that you don’t yet understand the definition of the quotient topology. – Brian M. Scott Oct 16 '13 at 02:55
  • I'll take another look at the quotient topology chapter. Thank you for your help though! – KangHoon You Oct 16 '13 at 03:30
  • @KangHoonYou: Probably a good idea. The quotient topology is one of the more difficult concepts in basic general topology, but it’s an important one. You’re welcome! – Brian M. Scott Oct 16 '13 at 03:35
  • @BrianM.Scott I just thought of one! How about f(x)={0 if irrationals>0, 1 if irrationals<0 and 2 of x is rational} – KangHoon You Oct 16 '13 at 03:53
  • @KangHoonYou: Yes, that will work just fine. – Brian M. Scott Oct 16 '13 at 03:59