Am I correctly identifying the fallacy in this induction “proof?”

The prompt states:

Let us accept as true that a person can always walk an extra mile. Does the Principle of Induction then prove that a person can walk forever? Where is the fallacy?

No. The Principle of Induction says that a person can walk a finite number of miles $n$ for all $n \in \mathbf{N}$, but does not say that a person can walk infinitely many miles (walking forever).

The fallacy is that Induction may be used to prove that a statement of the form $P(n)$ is true for all $n \in \mathbf{N}$. It cannot be used to show that a statement of the form $P(\infty)$ is true.

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There is no fallacy, it is possible to prove that a person can walk forever, under the assumption that he or she can always walk another mile. –  Git Gud Nov 17 '13 at 20:04
"Forever" does not have a clear meaning. If it means can walk more than $n$ for any $n$, then the implication is correct. But why should one accept that a person can always walk an extra mile? –  André Nicolas Nov 17 '13 at 20:06
@hardmath: you are playing with semantics here. "Walk forever" does not mean "walking infinitely many miles"; it means walking an arbitrarily large number of miles, given enough time. –  Martin Argerami Nov 17 '13 at 20:09
There's no fallacy. The assumption is false (that a person can always walk another mile). If a person could always walk another mile, a person could walk forever. –  mjqxxxx Nov 17 '13 at 21:01
Maybe part of the confusion comes from the fact that you talk about induction, but you don't state your proposition $P(n)$ explicitly. To me, here the natural $P(n)$ is $$P(n)=\mbox{"a person can walk n miles"}.$$ So the induction principle shows that a person can walk any number of miles. To me, this is what it means "to walk forever". Note that if a person "walks forever" at no time the person would have done infinitely many miles.
@hardmath A parallel can be drawn with the concept of a function being infinitely differentiable. Of course there does not exist such a thing as $f^{(\infty)}$, but it can happen that a function is such that $f^{(n)}$ exists for all $n\in \Bbb N$ and this is what we call an infinitely differentiable function. –  Git Gud Nov 17 '13 at 20:58