Difference between these assumptions?

Question

If given some statement $P$ about natural numbers, what is the difference between the following: "Assume $k \in \mathbb {N}$ and $P(k)$" $(1)$, and the second way, "Assume $\exists k \in \mathbb {N}: P(k)$"$(2)$. I have seen contradictory answers online and I would like full clarification, thanks.

Answer 1

I assume that you are bringing this up in the context of an inductive proof, whose inductive step usually starts out with something like

"Assume that $P(k)$ for some number $k$".

Now, the use of 'some' here makes it sound like your (2): $\exists k \in \mathbb {N} : P(k)$

However, this is certainly not the assumption of the inductive step. Consider: suppose we use that assumption. We could then also suppose that $k_0$ is one of those numbers, and maybe we can even show that that means that $k_0 +1$ has property $P$ as well. OK, so what have we then proven? All we would have proven is that there is some number such that it and its successor have property $P$. In logic: we would have shown that $\exists k \in \mathbb {N} : P(k) \land P(k+1)$. Or, discharging the assumption, we obtain $$\exists k \in \mathbb {N} : P(k) \to \exists k \in \mathbb {N} : P(k) \land P(k+1)$$ But that is not what the inductive step tries to show.

The inductive step tries to show that $\forall k \in \mathbb {N} : P(k) \to P(k+1)$. We do that by picking some arbitrary number, assume that it has property $P$, and use that to show that $P(k+1)$.

So here is the difference: instead of assuming $\exists k \in \mathbb {N} : P(k)$, we simply assume $P(k)$ once $k$ has been introduced as an arbitrary number. So, the assumption is really just $P(k)$. Once we obtain $P(k+1)$, we can discharge the assumption and get $P(k) \to P(k+1)$, and by noting that $k$ was an arbitrarily chosen number, we can generalize this for all numbers, and thus obtain $$\forall k \in \mathbb {N} : P(k) \to P(k+1)$$

So, the assumption is really more like your (1), where $k$ is a free variable. Indeed, when the inductive step starts out with:

"Assume that $P(k)$ for some number $k$"

you should really read that as:

"Assume that $P(k)$ for some arbitrary number $k$"

or, better yet, as:

"Let $k$ be some arbitrary number and assume that $P(k)$"

It is $k$ being an arbitrary number (as reflected by it being free) that allows you to conclude with the universal statement that you want.

Answer 2

"Assume $\exists k \in \mathbb{N} : P(k)$"

Assume there exists at least one natural number $k$ such that $P(k)$ holds true. The specific natural number for which $P$ is true is not named. You are merely told that one exists for which $P$ is true.

"Assume $k \in \mathbb{N}$ and $P(k)$."

$k$ is assumed to be a specific natural number for which $P(k)$ holds true.

Answer 3

"$\forall k {\in} \mathbb {N}\;P(k)$"

Here, $k$ is a bound variable pointing to an arbitrary natural number, which has property $P.$

"$\exists k {\in} \mathbb {N}\;P(k)$"

Here, $k$ is a bound variable pointing to an unspecified natural number that has property $P.$

"$k \in \mathbb {N}$ and $P(k)$"

Here, depending on the context, either:

1. $k$ is a particular natural-number constant that has property $P.$

or:

2. $k$ is a free natural-number variable that has property $P.$ Only in this case, the given formula is an open formula rather than a sentence, and has no definite truth value until $k$ is instantiated.

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"Assume that $P(k)$ for some natural number $k$."

As observed by Bram28, this sounds like a common phrasing in the inductive step of an induction proof. What's actually happening is that we are considering an arbitrary natural number $k,$ then assuming that $P(k)$ is true, then showing that $P(k+1)$ is consequently true; all in all, the inductive step is showing that

• $\color{green}{\textbf{for each}}$ natural number $k,$ if $P(k)$ is true, then $P(k+1)$ is true

  $\color{green}{\boldsymbol\forall} k{\in}\mathbb N\;\big(P(k)\implies P(k+1)\big);$

an equivalent phrasing is

• $\color{green}{\textbf{given some arbitrary}}$ natural number $k,$ if $P(k)$ is true, then $P(k+1)$ is true,

which unfortunately sounds like

• $\color{red}{\textbf{for some}}$ natural number $k,$ if $P(k)$ is true, then $P(k+1)$ is true

  $\color{red}{\boldsymbol\exists} k{\in}\mathbb N\;\big(P(k)\implies P(k+1)\big),$

understandably throwing people off.

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Reply to the OP's comment

Could you elaborate on the differences between the two "$k∈\mathbb N$ and $P(k)$" cases, and which one we use for induction?

The inductive step can be rewritten as $$\forall k \;\big(k\in\mathbb N\text{ and }P(k)\implies P(k+1)\big),$$ where $\text“k\in\mathbb N\text{ and }P(k)\text”$ in isolation corrresponds to the second case above.