I have a function $f$:

$$f:(0,\infty) \longrightarrow \mathbb R \\ x \mapsto\frac{1}{2}(e^{-x}+e^x)$$

I want to know what the domain of $f$ is. I know that $f(x)=\frac{1}{2}(e^{-x}+e^x)$ is valid for all $x \in \mathbb R$ so I was initially thinking that the domain is $\mathbb R$. However, my function only maps a subset of $\mathbb R$ namely $(0, \infty)$ to the real numbers so I wasn't sure if I need to take that into consideration when specifying the domain. I guess the more general question I asking is this:

Suppose I have a function: $$g:A \longrightarrow B \\ a \in A \mapsto g(a)=b \in B $$

Does the domain always have to be a subset of $A$ or can the domain have a "higher" cardinality than $A$.


The notation $g\colon A\longrightarrow B$ means (among other things) that the domain of $g$ is $A$ and not some other set.

And the cardinality of $\mathbb R$ is the same as that of $(0,\infty)$; it is not higher.

  • $\begingroup$ Thank you for your answer. I just realized I made a mistake when "comparing" the cardinality. Of course the are the same. Just a follow up question on the first part of your answer. Using my example: $f(x)=\frac{1}{2}(e^x+e^{-x})$, even though $f(x)$ is valid for all $x \in \mathbb R$ the fact that it was specified that $f$ maps $(0, \infty)$ to $ \mathbb R$ means that my domain is "only" $(0, \infty)$ then? $\endgroup$ – qmd Apr 27 '19 at 14:12
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    $\begingroup$ Yes, you are right about that. $\endgroup$ – José Carlos Santos Apr 27 '19 at 14:16
  • $\begingroup$ Thanks for your help. Have a nice day. $\endgroup$ – qmd Apr 27 '19 at 14:19

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