I know that this set is a subset of the set of all real functions. Hence its cardinality is less than or equal to $\aleph ^\aleph=2^\aleph$.

The question is how do I prove the second direction?

(actually I'm not sure that $2^\aleph $ is the cardinality of the above set)

  • $\begingroup$ HINT:This set contains the functions $3/x$ also $4/x$ and for every real number $c$ the function $c/x$ has this property. The bijection $c\mapsto f_c(x)=c/x$ is a bijection from $\Bbb R$ to a subset of the functions that aproaches to $0$ when $x\to\infty$. $\endgroup$
    – MphLee
    Jun 2, 2015 at 10:11
  • 1
    $\begingroup$ @mphLee that is true but doesn't that only give $2^{\aleph_0}$ many functions. Which will be strictly less then the amount there really are. $\endgroup$
    – DRF
    Jun 2, 2015 at 10:12
  • $\begingroup$ @DRF You're right...since it only shows that the answer can not be $\aleph_0$... but it doesn't say if the answer if $\mathfrak c$ or $2^\mathfrak c$. NOTEActually my hint is useless... it only gives a lower bound but not the anwser $\endgroup$
    – MphLee
    Jun 2, 2015 at 10:19
  • $\begingroup$ Are you asking about all real functions such that $\lim_{x\to\infty} f(x)=0$, or only continuous ones? $\endgroup$ Jun 2, 2015 at 11:22
  • $\begingroup$ @MartinSleziak Good point. That would rather change the answer. But the fact that the limit is finite at infinity still has no bearing on how many such functions there are. $\endgroup$
    – DRF
    Jun 2, 2015 at 11:38

1 Answer 1


First of all you are missing a subscript on that $\aleph$. Second of all the set of all functions from the reals to $2$ is going to be $2^{\mathfrak{c}}=2^{(2^{\aleph_0})}$. Where $\aleph_0=|\mathbb{N}|$.

Now in answer to your question the number of functions that converge to $0$ at infinity will be the maximum possible, that is the same as the number of functions from the reals to the reals without any restriction. You just need to think about how many functions from $[0,1]$ to $[0,1]$ there are. You can obviously extend any such function to have a limit of $0$ at infinity.


Just to explain a bit about what the $\aleph$-notation means so you can see why you need some subscripts. $\aleph$'s are infinite cardinalities in increasing size. The first such cardinality is denoted $\aleph_0$ and is the size of the least infinite cardinal which is actually generally denoted $\omega$. The second least infinite cardinal is $\omega_1$ which has cardinality $\aleph_1$. While very often in set theory people will not bother distinguishing $\aleph$'s and the corresponding $\omega$'s it's probably better to think of them as separate and whenever you're actually talking about the cardinality of a set you should the $\aleph$-notation (as you did) whereas whenever you are talking about the cardinal you should use the $\omega$-notation

One last note: I should probably point out what that strange $\mathfrak{c}$ is just in case. $\mathfrak{c}$ is the standard symbol used for the cardinality of the continuum or in other words the cardinality of the real numbers/powerset of the natural numbers. This might or might not be equal to $\aleph_1$, but some parts of set theory are somewhat boring if it is. Still in general the most "probable" choices for $\mathfrak{c}$ seem to be $\aleph_1$ (under the Continuum Hypothesis) or $\aleph_2$ (under the Proper Forcing Axiom) or possibly some unnamed largish cardinal if you want to play around with cardinal invariants of the continuum.

  • $\begingroup$ It should be noted that $\aleph$ is sometimes used as a notation for $\frak c$, the cardinal of the continuum. $\endgroup$
    – Asaf Karagila
    Jun 2, 2015 at 11:44
  • $\begingroup$ @AsafKaragila Really? Is that a Shelah school type convention (I'm thinking inverted; or correct depending on where you learned forcing; forcing conditions) or is it ubiquitous and I've just never seen it? $\endgroup$
    – DRF
    Jun 2, 2015 at 11:47
  • $\begingroup$ I think it goes back to Cantor. That's what I was told when I was a student. I don't think Shelah uses this particular notation, but I'll ask around. $\endgroup$
    – Asaf Karagila
    Jun 2, 2015 at 11:58
  • $\begingroup$ @AsafKaragila Thank you. Though it's certainly very possible that it's a less used but non-specific notation that I never came across. I wonder whether the OP (if he/she reads this) might say where he saw/learned it? $\endgroup$
    – DRF
    Jun 2, 2015 at 12:07
  • $\begingroup$ You can find it in relatively elementary questions (like this one) across the site. In most of these questions there is a comment like the one I made here (often made by me). $\endgroup$
    – Asaf Karagila
    Jun 2, 2015 at 13:17

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