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Prove that Real Numbers and the interval $(0,\infty)$ have the same cardinality.

Attempt:

Consider the function $f(x) = e^x$.

The domain of this function is all real numbers.

The range of this function is from $0$ to infinity.

Let $e^a = e^b$.

Then $\ln(e^a) = \ln(e^b)$

Then $a\ln(e) = b\ln(e)$

This means that $a = b$

Hence, $f$ is injective.

Let $c > 0$

Then $e^{ln(c)} = c$

Since $c > 0, \ln(c)$ is defined, so $f(\ln(c)) = c$

Therefore, f is surjective.

Then f is bijective.

Hence, $\mathbb{R}$ and $(0, \infty)$ have the same cardinality.

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    $\begingroup$ What is your question? If your question is "is my proof correct?", then the answer is "yes": the function $x \mapsto e^ x$ is indeed a bijection between $\mathbb{R}$ and the interval $(0, \infty)$. Hence these two sets have the same cardinality. $\endgroup$ – Rob Arthan Nov 19 '13 at 21:59
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This is a valid proof. Essentially you have used the fact that the exponential function has a two-sided inverse (at least when the codomain is the positive reals), $\log$ (or $\ln$ if you prefer). Any function with a two-sided inverse is automatically a bijection. In fact, injectivity is precisely the same as having a left inverse and surjectivity is precisely the same as having a right inverse. Exercise: if a function $g$ has a left inverse $f$ and a right inverse $h$, then in fact $f = h$, so $g$ has a two-sided inverse.

Here are some other approaches you could use:

  • observe that the exponential function has positive derivative everywhere, so it is increasing, so it must be injective (roughly speaking, use the contrapositive of Rolle's theorem).
  • show (somehow, not sure what you're allowed to assume) that the exponential function takes arbitrarily small (positive) values, and arbitrarily big values, so by the Intermediate Value Theorem takes all the values in between, so must be surjective.
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    $\begingroup$ Indeed; the hard part of the proof has been done when you assume the existence of an inverse function $\ln(x)$. The above proof can basically be shortened to: Consider the function $f(x) = e^x$. This is known to be a bijection from $\mathbb{R} \to (0, \infty)$. Therefore, $\mathbb{R}$ and $(0, \infty)$ have the same cardinality. $\endgroup$ – augurar Jul 24 '15 at 20:12

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