Induction is normally defined for the case when you have an explicit dependence on an integer (or an ordinal for transfinite induction, but I'm not going to talk about that). I'm not sure what you're talking about with the function, but since induction requires you to prove that $P(k)\Longrightarrow P(k+1)$ for *all* $n$, you wouldn't be able to 'prove' that the function was everywhere positive, because the statement $P\Longrightarrow P(n+1)$ would fail at the point where we cross the 'limit point'.

You can prove that proof by induction is a proof as follows:

Suppose we have that $P(1)$ is true, and $P(k)\Longrightarrow P(k+1)$ for all $n\ge 1$. Then suppose for a contradiction that there exists some $m$ such that $P(m)$ is false. Let $S=\{n\in \mathbb{N} : P(k) \text{ is false}\}$. $S$ is non-empty (since it contains $m$), so it has a least element $s$ (The statement that every non-empty set of natural numbers has a least element is known as the **Well Ordering Principle**).

Now, since $P(1)$ is true, $s\neq 1$. So $s-1$ is a natural number. Now, if $s-1$ were true, then $P(k)\Longrightarrow P(k+1)$ would imply that $s$ were true (setting $k=s-1$). Since $s$ is not true, $s-1$ must not be true as well. But $s-1<s$, and we had supposed $s$ to be the *smallest* natural number $n$ such that $P(n)$ was not true. So we have a contradiction.

Therefore, $P(n)$ is true for all $n\ge 1$.

It turns out there actually is an analogue of induction which you can use if you want to prove that a statement is true for all real numbers. You can't apply the above argument to the real numbers, as the real numbers don't satisfy the Well Ordering Theorem. For example, the set $\{1,0.1,0.01,0.001,\dots\}$ doesn't have a smallest element. Neither does $\{x\in \mathbb{R} : x>3\}$. However, the real numbers do satisfy what is known as the **Greatest Lower Bound property**. This means that every set of real numbers which is bounded above has a *greatest lower bound*. For example, the set of real numbers that are greater than $3$ has no least element: if you take some number greater than $3$ - $3.001$, say - I can tell you another number that is greater than $4$ but still less than your number (like $3.0001$). But it does have a greatest lower bound, which is $3$. The set $\{x\in\mathbb{R} : x^2 > 2\}$ has no least element, but it has a greatest lower bound, which is $\sqrt{2}$.

**Exercise:** what is the greatest lower bound for the set $\{1,0.1,0.01,0.001,\dots\}$?

In general, a lower bound for a set is a number $b$ that is smaller than every element in the set. The Greatest Lower Bound Property is the statement that whenever you've got a lower bound, there is always a *greatest* lowers bound. We denote the greatest lower bound of a set $S$ by $\inf S$ ($\inf$ is short for *infimum*, another word for *greatest lower bound*).

How can we use the Greatest Lower Bound Property to form an analogue of induction? I'll give you the statement of continuous induction first, and then prove it.

**Theorem** Let $P(x)$ be a statement about an arbitrary real number $x$. Suppose we know the following two facts:

- $P(x)$ is always true if $x<y$, where $y$ is a fixed real number.
- If $P(x)$ is always true up to some number $c$ (whenever $x<c$) then, for some $\varepsilon > 0$, $P(x)$ is always true up to $c+\varepsilon$ (whenever $x<c+\varepsilon$).

Then $P(x)$ is true for all real numbers $x$.

The proof is similar to the proof for natural-number induction:

**Proof:** Suppose for a contradiction that, for some real number $z$, $P(z)$ is false. Let $A=\{x\in \mathbb{R} : P(x) \text{ is false}\}$. Since $P(x)$ is true for all $x<y$, $y-1$ is a lower bound for $A$. Since $z\in A$, $A$ is non-empty. So $A$ has a greatest lower bound $a=\inf A$. Now, if $x<a$, then $P(x)$ is true (as $a$ is a lower bound for $A$, the set of real numbers $x$ such that $P(x)$ is false). Therefore, for some $\varepsilon > 0$, $P(x)$ is true for all $x<a+\varepsilon$. Therefore, $a+\varepsilon$ is a lower bound for $A$. But $a+\varepsilon>a$, and we had supposed $a$ to be the *greatest* lower bound for $A$. So we have a contradiction.

So $P(x)$ is true for all real numbers $x$.

This is really the same thing as proof by induction for integers, but using the Greatest Lower Bound Property instead of the Well Ordering Principle. It is worth mentioning that I have *never* had occasion to use continuous induction over real numbers, and it's not a standard technique: mathematicians tend to prove statements about the real numbers straight from the Greatest Lower Bound Principle, and not use any fancy tricks (just as you can turn any proof by induction into a proof where you try to find a 'minimal counterexample', as I did when I proved that induction works above). But it's quite fun, and it shows that you can generalize things like the induction principle on to more general sets, as long as you have some statement similar to the Well Ordering Principle.