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

  • 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

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.

  • $\begingroup$ 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

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