Peano’s Fifth axiom as stated by Russell

Question

I’m translating Russell’s Introduction to Mathematical Philosophy. I’m having difficulty understanding his formulation of the fifth axiom:

(5) Any property which belongs to 0, and also to the successor of every number which has the property, belongs to all numbers.

Especially, I don’t understand the property that belongs to zero and to all successors? What is that property?

I know the Fifth axiom can be stated different ways as in this question but can you answer according to Russell’s wording?

And why is this considered a definition of mathematical induction?

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Edit

This is an edit regarding Ann Bauval's comment on December 12 to be on the same page about notational conventions.

I understand from comments that there are three different conventions to write the indices. I'll write each explicitly for $n=3$.

(1) The first is your initial version with index written as $0 \leq k < n$. In this case $k_{final} = n-1$ and $k$ can be zero or greater than zero.

$\displaystyle\sum_{0\leq k < n} 2^k = 2^n - 1$

For $n=3$,

$2^0 + 2^1 + 2^2 = 2^3 - 1$

(2) The second version is the one I mentioned on my comment on December 3:

$\displaystyle\sum_{k=0} ^{n} 2^k = 2^{n+1} - 1$

In this case $k$ goes to $n$, inclusive:

$2^0 + 2^1 + 2^2 + 2^3 = 2^4 - 1$

And for $n=0$ (for induction test):

$2^0 = 2^1 - 1$

and the induction test for $n=0$ is satisfied.

(3) Then today, December 12, you introduced a new notation as $n-1$ for the top index:

$\displaystyle\sum_{k=0} ^{n-1} 2^k = 2^{n} - 1$

I don't see the point of introducing such final index. For $n=0$ and $k=0$ we would be saying "sum $2^k$ from 0 to 0" which is a meaningless expression. From this it is clear that we need to state as an initial condition that $n>k$.

Please let me know if I misunderstand any of these notational conventions.

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Edit

Edit about Anne Bauval's comment (11 December).

I wrote the expansions for $\sum_{k=0} ^{n} 2^k = 2^{n+1} - 1$ from $n=3$ to $n=0$:

\begin{align*} 2^0 &+ 2^1 + 2^2 + 2^3 &= 2^4 - 2^0 \quad \quad (n=3)\\ 2^0 &+ 2^1 + 2^2 &= 2^3 - 2^0 \quad \quad (n=2)\\ 2^0 &+ 2^1 &= 2^2 - 2^0 \quad \quad (n=1)\\ 2^0 & &= 2^1 - 2^0 \quad \quad (n=0) \end{align*}

Answer 1

It is simply the original (1889) Peano's axiom of induction:

"if $K$ is a class such that: $0$ is in $K$, and for every natural number $n$, $n$ being in $K$ implies that the successor of $n$ is in $K$, then $K$ contains every natural number"

rewritten using property instead of class (i.e. set).

This matches quite exactly with the modern version you have linked above:

If a subset $T$ contains $0$ as an element and for all $n \in T$, $s(n)$ [i.e. the sucecssor of $n$] is also in $T$, then $T$ is the set of Natural numbers $\mathbb N$.

A "property" is e.g. "being Even"; thus to say that a number $n$ has property $E(x)$, i.e. that $E(n)$ holds, is equivalent to say that $n \in E$ where $E$ is the set of even numbers.

The intuitive understanding of properties and classes is that every property "identifies" a class, i.e. the class of all and only those objects that satisfy the property.