Is it correct to define the positive real numbers as $\{f(x) = x^2\mid x \in \mathbb R\}$?
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2$\begingroup$ You probably mean $\{x^2\mid x\in\mathbb{R}\}$. That is a description of the set of non-negative reals. $\endgroup$– André NicolasCommented Oct 21, 2014 at 3:12
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1$\begingroup$ The image of $f(x)=\dfrac{1}{|x|}$ will give you all positive reals. $\endgroup$– David PCommented Oct 21, 2014 at 3:16
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1$\begingroup$ @David: Of course this wasn't mentioned in the question, but if you're going to start nitpicking and remove points from the domain of the function, why not the identity function on the positive reals and that's it? $\endgroup$– Asaf Karagila ♦Commented Oct 21, 2014 at 3:20
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3 Answers
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Why don't use such that $\{x\in \Bbb R: x>0\}$?
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$$ \{y^2:y \in \mathbb{R} \backslash \{x^2:x \in \mathbb{R}\}\} $$
answered Oct 21, 2014 at 3:21
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$\begingroup$ Maybe try to explain why this is an answer to the question? $\endgroup$ Commented Oct 21, 2014 at 4:31
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1$\begingroup$ @David-Holden This is clever. $\endgroup$ Commented Oct 21, 2014 at 4:49
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Try the exponential function. Paul, he wanted to do it using a function.
answered Oct 21, 2014 at 3:22
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$\begingroup$ Why is this half a comment? Comments go to the comments, answers go to the answers. $\endgroup$– Asaf Karagila ♦Commented Oct 21, 2014 at 3:24
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$\begingroup$ Also, it's not clear that E wanted to do it using a function. Here is a reason why E might have not: $x^2$ has the advantage over $e^x$ that it is defined completely algebraically, showing that we can build the topology from the algebraic structure. $\endgroup$ Commented Oct 21, 2014 at 4:32