Do $|\mathbb{Z}^{+}| < |\mathcal{P}(\mathbb{Z}^{+})|$ and $|\mathcal{P}(\mathbb{Z}^{+})| \leq |\mathbb{R}|$ imply $|\mathbb{Z}^{+}| < |\mathbb{R}|$?

Question

I am reading "A First Course in Analysis vol.3" written by Kazuo Matsuzaka.

First, he proves $\lvert X \rvert < \lvert \mathcal{P}(X) \rvert$ for any set $X$.

Next, he proves $\lvert \mathcal{P}(\mathbb{Z}^{+}) \rvert \leq \lvert \mathbb{R} \rvert$.

Then, he concludes that $\lvert \mathbb{Z}^{+} \rvert < \lvert \mathbb{R} \rvert$ because $\lvert \mathbb{Z}^{+} \rvert < \lvert \mathcal{P}(\mathbb{Z}^{+}) \rvert$ and $\lvert \mathcal{P}(\mathbb{Z}^{+}) \rvert \leq \lvert \mathbb{R} \rvert$.

But I cannot understand why $\lvert \mathbb{Z}^{+} \rvert < \lvert \mathbb{R} \rvert$ holds.

Answer 1

Generalising slightly: for arbitrary sets $X$, $Y$ and $Z$, to say that $|X| < |Y| \le |Z|$ means that there are injections $f : X \to Y$ and $g : Y \to Z$ but there is no injection $Y \to X$. Let's prove that this implies that $|X| < |Z|$.

To deduce that $|X| \le |Z|$, note that $g \circ f : X \to Z$ is injective.

To finally deduce that $|X| < |Z|$, suppose there were an injection $h : Z \to X$. Then the composite $h \circ g : Y \to X$ would be injective, contradicting the assumption that $|X| < |Y|$. So indeed we have $|X|<|Z|$.

So here, let $X=\mathbb{Z}^+$, $Y=\mathcal{P}(\mathbb{Z}^+)$ and $Z=\mathbb{R}$ and apply the above result.

Answer 2

The relations $<$ and $\le$ are both transitive. Thus, if we have, for cardinals $A,B,C$, that $A<B<C$, then we can conclude $A<C$. Similarly, if we have $A\le B\le C$, we can conclude $A\le C$. This is a combination of the two, namely: $A<B\le C$.

Since $A<B\implies A\le B$, then we can plug that in, and obtain $A\le C$. Then, if we can rule out $A=C$, we will have $A<C$. However, $A=C$ is impossible, for we know $B\le C$, and substituting $A$ for $C$ contradicts what we know about $A$ and $B$.

In this particular case, we have $A=|\Bbb Z^+|$, $B=|\mathcal P(\Bbb Z^+)|$, and $C=|\Bbb R|$ Does this answer your question?