1
$\begingroup$

Try:

Call $M$ to be the set of all maximum points of a given $f: \mathbb{R} \to \mathbb{R}$. We wish to construct $\phi: \mathbb{N} \to M $ so that $\phi$ is bijective.

Notice that $x_i$ is maximum of $f$ is we can find $\epsilon_i$ so that ....

Maybe we can define $\phi(i) = x_i $ and check for bijectivity.

If $\phi(i) = \phi(k) \implies x_i = x_k \implies i=k $

Now, any points $x_j$ where $f$ attains its maximum there corresponds an index $j$ so that surjectivity follows and we have bijectivity.

Thus, $M$ is finite our countable.

Is my approach correct?

$\endgroup$
  • $\begingroup$ Note that the definition allows you to associate to each maximum point an open interval having a certain property such that this property implies that the open intervals are pairwise disjoint. Thus, it is enough to prove that any collection of pairwise disjoint open intervals is at most countable, a result that you might already have available, but if not, think about what you can deduce if you pick a rational number from each of these open intervals (you'll get at most countably many rationals, no two of which belong to the same open interval . . .). $\endgroup$ – Dave L. Renfro Jan 24 at 9:02
1
$\begingroup$

No. When you say that $x_i$ is a maximum point, you're assuming that such points are countable.

You can start by proving that the each maximum point is isolated and then use the fact that discrete subsets of $\mathbb{R}$ are at most countable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.