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Set $A \subset R^2$, set B is projection of A on x-axis. Do you know a counterexample to the statement: if A is closed, then B is closed.

John
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2 Answers2

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Hint: Consider the graph of $\tan(x)$ between two asymptotes.

Asaf Karagila
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    Interesting example. Just to make sure that I got it right: any continuous line (like tan betwenn $\frac {-\pi}{2}$ and $\frac {\pi}{2}$) is closed because it contains all its limit points. Also, any point of the continuous line is limit point. right? – John Jul 04 '13 at 15:41
  • @John: True (though I'd say "curve" instead of "line")! Nice non-canonical counterexample, Asaf! – Cameron Buie Jul 04 '13 at 15:42
  • @Cameron: Thanks. I just happened to think about that one a few days ago (as a nice solution for this very problem). I actually thought about "The graph of any homeomorphism from $(0,1)$ to $\Bbb R$", but the principle is the same. – Asaf Karagila Jul 04 '13 at 15:43
  • Thank you, Asaf! It's always funny when the example OP needs is quite obvious after being told about it. – John Jul 04 '13 at 15:46
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    I always contented myself with ${xy=1}$. That was lazy, +1. – Julien Jul 04 '13 at 15:53
  • @julien: Strange how when I thought about it a few days ago, it didn't pop right into my head... :-) – Asaf Karagila Jul 04 '13 at 15:58
  • I think it is a bit like I would automatically take $\sin(1/x)$ to produce a connected non path-connected space. – Julien Jul 04 '13 at 16:01
  • @julien: We would? I recall some sort of comb space being the example... – Asaf Karagila Jul 04 '13 at 16:03
  • Edited: I would... – Julien Jul 04 '13 at 16:04
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    @julien: I would eat a piece of wood if I had to carry out the $\log$! – Asaf Karagila Jul 04 '13 at 16:05
  • I disagree with @Git Gud: I am pretty sure you are the funniest person on MSE. – Julien Jul 04 '13 at 16:18
  • @julien: Hah! In the sixth grade I was the least funny kid in my class. I still have a great tendency to make tons of corny jokes (from a recent conference, they were discussing how come there is no Cantor Research Center, "Is there a Turing Research Center For Computational Sciences or something? I'm sure many people would love to have a Turing degree!" - the table goes silent. And that was after the first bottle of wine!) – Asaf Karagila Jul 04 '13 at 16:21
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Take $A = \{(x,y) \in \Bbb{R}^2 \hspace{2mm} | \hspace{2mm} xy = 1\}$ and $B$ being the projection of $A$ onto the $x$ - axis. Then $A$ is closed being the zero set of the polynomial $xy - 1$ but $B$ is not as it is $\Bbb{R} - \{0\}$.