Show that set of real numbers is equipotent to the set of positive real numbers.

Question

1) Show that R ≈ R+.
2) Help me to show that set of real numbers is equipotent to set of positive real numbers.

epsilon · Answer 1 · 2013-12-11T15:30:35.450

2

The function $f(x)=\log x$ is one-to-one and onto from $\{x\in\mathbb{R}|x>0\}$ to $\mathbb{R}.$

edited Dec 11 '13 at 15:30

answered Dec 11 '13 at 15:07

epsilon

1,786

What's the image of $0$? – Jean-Claude Arbaut Dec 11 '13 at 15:08
1

$0$ does not belong to the domain of $f$. – epsilon Dec 11 '13 at 15:09
Yes, so it's not a bijection between $\Bbb R_+$ and $\Bbb R$. – Jean-Claude Arbaut Dec 11 '13 at 15:10
Does $0\in \mathbb{R}^+$? – epsilon Dec 11 '13 at 15:11
Yes, at least in my interpretation of it. For me, $\Bbb R_+ \setminus {0} = \Bbb R_+^*$. But the OP says "positive numbers", so I guess you are right. It's a known divergence between french and english: in french, positive usually mean $\geq 0$, and $>0$ is "strictly positive", while in english it's nonnegative for $\geq 0$ and positive for $>0$. However, concerning set notation, I believe the one I use is quite usual. – Jean-Claude Arbaut Dec 11 '13 at 15:12
Well... I guess even this notation is not that usual: http://math.stackexchange.com/questions/27968/how-does-one-denote-the-set-of-all-positive-real-numbers – Jean-Claude Arbaut Dec 11 '13 at 15:21
I'll edit my post to avoid confusion, thanks. – epsilon Dec 11 '13 at 15:29

score 2 · Answer 2 · answered Dec 11 '13 at 15:35

If you want a bijection from $\mathbf{R}$ to $\{x \in \mathbf{R} : x \geq 0\}$, first notice that the function $f:x \mapsto e^x$ is a bijection from $\mathbf{R}$ to $\{x \in \mathbf{R} : x > 0\}$, the inverse is obviously the logarithm. However, we alter our function $f$ by sending $1 \mapsto 0, 2 \mapsto e^1, 3 \mapsto e^2, \dots, k \mapsto e^{k-1}, \dots.$ The resulting function is bijection from $\mathbf{R}$ to $\{x \in \mathbf{R} : x \geq 0\}$ as desired.

Jean-Claude Arbaut · Answer 3 · 2013-12-11T15:23:01.383

I assume you mean $\Bbb R_+ = \{ x \in \Bbb R \;|\; x \geq 0\}$. If you mean $\Bbb R_+ = \{ x \in \Bbb R\;|\; x > 0\}$, then it's what I denote $\Bbb R_+^*$, and you can skip the following sentence.

You don't change cardinality of an infinite set by adding or removing finitely many elements. Hence, it's enough to prove $\Bbb R \simeq \Bbb R_+^*$.

Now it's fairly easy to find a bijection between these two sets, take for example $x \to e^x$.

Show that set of real numbers is equipotent to the set of positive real numbers.

3 Answers3

Linked