show that for an infinite cardinal $k$, $k + k = k$

Question

Show that for an infinite cardinal $k$, $k + k = k$

So far I have that $k + k = 2k$

Is it possible to somehow show that $2k = k$?

I've been trying to understand some cardinal arithmetic, and I know that if you have two disjoint sets $S,T$ with $|S| = m$ and $|T| = n$ then $ mn = |S \times T|$

but I don't know what do do when it's just an integer.

Answer 1

Well. I couldn't quite find a duplicate, so I'll write an answer instead. I will sketch the idea, and give some of the details. The rest I leave to you.

First we need to understand what $2k=k$ means. It means that if $K$ is a set such that $|K|=k$, then $\{0,1\}\times K$ and $K$ have the same cardinality. You can easily see that $\{0,1\}\times K=\{0\}\times K\cup\{1\}\times K$, and this is the disjoint union of two sets of cardinality $k$.

For example, $\aleph_0+\aleph_0=\aleph_0$ (where $\aleph_0=|\Bbb N|$).

Next we are going to need the following fact: $K$ is infinite if and only if there is a countably infinite subset of $K$. Or in other words, $\aleph_0\leq k$.

As well one of its consequences, if $K$ is infinite and $A$ is finite, then $|K\setminus A|=|K\cup A|=|K|$.

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Now we can use Zorn's lemma, which states that every partial order $(P,\leq)$ in which every chain has an upper bound has a maximal element.

We define the partial order on pairs of the form $(A,f)$ where $A\subseteq K$ is infinite and $f$ is a bijection from $\{0,1\}\times A$ onto $A$. Since $K$ is infinite it has countably infinite subsets, all of which has such $f$'s, so this partial order is certainly not empty.

We say that $(A,f)\leq(B,g)$ if $A\subseteq B$ and $f\subseteq g$, meaning $A$ is a subset of $B$ and if we restrict $g$ to $\{0,1\}\times A$ we get $f$.

Finally, if $\{(A_i,f_i)\mid i\in I\}$ is a chain, then I claim that $A=\bigcup A_i$ and $f=\bigcup f_i$ are such that $(A,f)$ is in the partial order. Certainly $A$ is infinite. Let us see that $f\colon\{0,1\}\times A\to A$ is a bijection:

1. First of all, $f$ is a function since it is the increasing union of functions.

2. For the same reason it is injective. If we have $t,t'\in\{0,1\}$ and $a,a'\in A$ and suppose that $f(t,a)=f(t',a')$. There is some $i\in I$ where $a,a'\in A_i$ and therefore $f_i(t,a)=f(t,a)=f(t',a')=f_i(t',a')$, but $f_i$ is injective so we have that $t=t'$ and $a=a'$.

3. And again this is a surjection, since if $a\in A$ there is some $i$ such that $a\in A_i$, and I will let you finish the argument from here.

Now, to finish the proof you have to show that the maximal element is $(K,f)$. But this might not be true. You can prove, however, that if $(K',f')$ is a maximal element then $K\setminus K'$ is finite. If it were infinite, you could have found a way to extend it beyond its maximality.

So we return to the consequence of the infinitude of $K$, and we have that $|K|=|K'|$ and therefore $2k=k$.