0
$\begingroup$

The question given is check Whether following is countable or not

$1).$The set of all the functions from $\mathbb{Q}$ to $\left \{0,1 \right \}$

$2).$The set of all the functions from $\mathbb{Q}$ to $\left \{0,1 \right \}$ which vanish outside a finite set

solution i tried- In $1$ option we can see that number of functions is $2^{\aleph_0}$ which is uncountable

But for $2$ option i am not getting clue how to proceed further Seems like he is asking about the cardinality of set of compact support functions .

Please help

Thank you

$\endgroup$
4
  • 1
    $\begingroup$ It sounds like you might be misinterpreting (2) as referring to functions with support contained in an interval of finite measure. Such a support can still be infinite, but the intended meaning of (2) is actually that $\{x\mid f(x)=1\}$ is a finite set. $\endgroup$ – Karl Apr 21 '20 at 16:59
  • $\begingroup$ but we have to find the cardinality of set with that function but in your comment it is cardinality of set of elements which maps to $1$,can you please elaborate? $\endgroup$ – honey kumar Apr 21 '20 at 17:10
  • $\begingroup$ I wasn't answering (2), just clarifying that you're not meant to count any function that's 1 on infinitely many rational numbers, even if all of those numbers are inside an interval of finite length. $\endgroup$ – Karl Apr 21 '20 at 17:15
  • $\begingroup$ In other words, don't worry about how the elements of $\Bbb Q$ are arranged on the number line. Just think of $\Bbb Q$ as $\Bbb N$, since $\Bbb Q$ is countable. You can interpret (2) as asking for the cardinality of the collection of all finite subsets of $\Bbb N$. $\endgroup$ – Karl Apr 21 '20 at 17:29
2
$\begingroup$

For each $n$, show that the set of functions from $\mathbb{Q}$ to $\{0,1\}$ that vanish outside a finite set of $n$ elements is countable. From here can you conclude that your set is countable?

For example, when $n=1$, we're looking for the functions that vanish at all but one point $x\in \mathbb{Q}$. Then we can easily see that there are two varieties of a such functions: (1) $f(y)=0$ for all $y\in\mathbb{Q}$, or (2) $f(y)=0$ for all $y\neq x$ and $f(x)=1$. Thus there are two functions for each $x\in \mathbb{Q}$, so this set is countable.

$\endgroup$
5
  • $\begingroup$ what is $y$ here? $\endgroup$ – honey kumar Apr 21 '20 at 17:06
  • $\begingroup$ $x$ was a distinguished element of $\mathbb{Q}$ (the unique element of the finite set in question). I used $y$ to refer to a generic element of $\mathbb{Q}$. $\endgroup$ – Nathan Lowry Apr 21 '20 at 17:26
  • $\begingroup$ i am still not getting this ,if each $x$ has two functions and $x \in \mathbb{Q}$ then still there are uncountable functions.Please can you elaborate your answer $\endgroup$ – honey kumar Apr 24 '20 at 8:25
  • $\begingroup$ For $n=1$, the functions are in bijective correspondence with the set $\mathbb{Q}\times \{0,1\}.$ A finite product of countable sets is countable. $\endgroup$ – Nathan Lowry Apr 24 '20 at 13:48
  • $\begingroup$ Maybe here's a more constructive hint: Show that for each $k$, the functions of this kind are in bijective correspondence with $\mathbb{Q}^k \times \{0,1,\dots, 2^k\}$. $\endgroup$ – Nathan Lowry Apr 24 '20 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.