You're welcome, happy to help!

[Joppy]

Loosely speaking, we say that periodic orbits are 'dense' if given any $\epsilon$-neighborhood, there exists at least one periodic point in that neighborhood for any $\epsilon > 0$.

Is there any requirement for these periodic points to be unique?

For example, what if every neighborhood contains a periodic point (that we know about) which is part of the same periodic orbit. Do we still say that orbits are dense? Or are they dense in a trivial sense.

Thanks!

 

[S.G. Janssens]

Loosely speaking, we say that periodic orbits are 'dense' if given any $\epsilon$-neighborhood, there exists at least one periodic point in that neighborhood for any $\epsilon > 0$.

For concreteness, is your setting a discrete dynamical system represented by an iterated map $f$ on $\mathbb{R}^n$ or, more generally, on some complete metric space? It does not matter that much, though: For a flow, the ideas are similar.

Yes, periodic orbits of $f$ are dense if given any point $\mathbf{x} \in \mathbb{R}^n$ and any $\epsilon > 0$, there exists a periodic point $\mathbf{y} \in \mathbb{R}^n$ of $f$ (of course $\mathbf{y}$ lies on some periodic orbit) such that $\|\mathbf{x} - \mathbf{y}\| < \epsilon$. In other words, the set consisting of the union of all periodic orbits of $f$ is dense in $\mathbb{R}^n$ in the ordinary sense.

Is there any requirement for these periodic points to be unique?

No, quite the opposite. Let $\mathbf{x}$ and $\epsilon > 0$ be given. Assume that $\mathbf{x}$ itself is not periodic. (This is always possible, unless the whole phase space consists of periodic points.) Suppose it happens that $\mathbf{y}$ is the unique periodic point in the $\epsilon$-ball centered at $\mathbf{x}$. Then the ball centered at $\mathbf{x}$ with radius $\frac{1}{2}\|\mathbf{x} - \mathbf{y}\| > 0$ does not contain any periodic points, which contradicts density.

For example, what if every neighborhood contains a periodic point (that we know about) which is part of the same periodic orbit. Do we still say that orbits are dense? Or are they dense in a trivial sense.

Sure we still say that periodic orbits are dense. In fact, the property you mention now is stronger than what you mentioned at the beginning: Now, one periodic orbit has to do the job of being dense in the phase space, whereas before it was only required that all periodic orbits together form a dense union.

Addition: Note that for the case of $\mathbb{R}^n$, a single periodic orbit cannot be dense. (In the general metric case, it is different.)

 

[Joppy]

Thanks a lot for the informative response, and apologies for my late one!

For concreteness, is your setting a discrete dynamical system represented by an iterated map $f$ on $\mathbb{R}^n$ or, more generally, on some complete metric space? It does not matter that much, though: For a flow, the ideas are similar.

Not quite $\mathbb{R}^n$, although I was assuming this in my question to make sure I see the other half of the story..

No, quite the opposite. Let $\mathbf{x}$ and $\epsilon > 0$ be given. Assume that $\mathbf{x}$ itself is not periodic. (This is always possible, unless the whole phase space consists of periodic points.) Suppose it happens that $\mathbf{y}$ is the unique periodic point in the $\epsilon$-ball centered at $\mathbf{x}$. Then the ball centered at $\mathbf{x}$ with radius $\frac{1}{2}\|\mathbf{x} - \mathbf{y}\| > 0$ does not contain any periodic points, which contradicts density.

Sure we still say that periodic orbits are dense. In fact, the property you mention now is stronger than what you mentioned at the beginning: Now, one periodic orbit has to do the job of being dense in the phase space, whereas before it was only required that all periodic orbits together form a dense union.

Addition: Note that for the case of $\mathbb{R}^n$, a single periodic orbit cannot be dense. (In the general metric case, it is different.)

This makes sense! An easy to understand argument indeed. Thanks.