Fdtxjr

Find the Cardinality of the set of all finite subsets of $mathbbR$.

I have proved that the set of all finite subsets of $mathbbN$ is countable . But I cannot find the cardinality of the set in case of $mathbbR$ .

My Attempt:

First I have considered the set $$A_k= a_i in mathbbR and a_1<...<a_k $$

Then $S=The set of all finite subsets of mathbbR=bigcup_n=1^inftyA_n$

and Now in the case of $mathbbN$ I could show that this $A_k$ is countable and consequently the set of all finite subset of $mathbbN$ is countable.

But in this case I cannot say anything like this...In this case the set will be of the cardinality as that of $mathbbR$ (I think) ..But cannot prove it...

Please Help.

We can show that $#S ge c$ via the injection $f: mathbbR rightarrow S$, $f(x) = x$.

If you show $#S le c$ then you can conclude $#S = c$ invoking Bernstein's theorem.

Let's prove the following: let $X = X_k, k in mathbbN$ be a family of subsets such that $#X_i le c$, $ forall i in mathbbN$, and let $V = bigcup_kin mathbbNX_k$. Then $#V le c$.

Because $#X_i le c$, we have for each $i$ a surjective $f_i: mathbbR rightarrow X_i$. We define $g: mathbbN times mathbbR rightarrow V$, $g(n,x) = f_n(x)$.

$g$ is an surjective function: let $x in V$, then $x in X_i$ for some $i$, then we have $a in mathbbR$ such that $f_i(a) = x$ because $f_i$ is surjective. Then $g(i,a) = f_i(a) = x$, with $(i,a) in mathbbNtimesmathbbR$.

We conclude that $g$ is surjective.

Thus $#V le #(mathbbNtimesmathbbR)$, this is, $#V le c$.

Your set $S$ satisfies the hypotheses, so we have $#S le c$, and thus $#S = c$.