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Let $S$ be a dynamical system on a metric space $X$ with discrete time $\mathbb{N}_0$.

In our script we have a theorem that says one can extend such a system to one, here called $\tilde{S}$, with continuous time $[0,\infty)$ on a larger space $Y$ and $\tilde{S}(1)|_X=S(1)$.

We don't have a proof, only the hint to use $Y=C^0([0,1])$. Can anyone help me with a proof?

Rino
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You construct an equivalence relation on $X\times [0,1]$ so that $(x,1)$ is equivalent to $(f(x),0)$ where $f=S(1,\cdot)$ is the step-1 map of the dynamical system $S$. Then define $$\tilde S(t,(x,s))=(f^n(x),\alpha)$$ where $s+t=n+\alpha$, $n\in\Bbb N_0$ and $\alpha\in[0,1)$.

Adapt to your notation convention if I guessed the interpretation of your symbols wrong.

Justification: The task description tells that $\tilde S$ acts on a bigger space $\tilde X$ where $X$ is embedded, $\iota:X\to\tilde X$. What we originally know with certainty is that the dynamic on the embedded set should be inherited, $$\tilde S(n,\iota(x))=\iota(S(n,x))=\iota(f^n(x)).$$

What this construction does is to find the most trivial space that satisfies the demands of the task, essentially turning the problem description into a solution.

So a value of $\tilde S(t,\iota(x))$ is needed for $t\in(0,1)$. The easiest construction is to give the pair $(x,t)\in X\times\Bbb R$ as that value. Next one has to ensure continuity at $t=1$, $$ \lim_{t\to 1}\tilde S(t,\iota(x))=\tilde S(1,\iota(x))=\iota(f(x)). $$ This leads directly to the construction of an equivalence relation $(x,1)\sim (f(x),0)$, easily generalized to $(x,t+n)\sim (f^n(x),t)$. Then $\tilde X=(X\times\Bbb R)/\sim$, the topology is inherited from $X\times\Bbb R$, the continuity of $\tilde S$ in all arguments follows from the continuity of $f$.

Lutz Lehmann
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  • Thank you @Lutz. I don't understand how to completely define this equivalence relation and what purpose it serves ... Could you explain it? And how do I get to the proposed space Y that was given as a hint? – Rino Jun 08 '20 at 14:27
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    Your task description tells that $\tilde S$ acts on a bigger space where $X$ is embedded. What this construction does is to find the most trivial space that satisfies the demands of the task, essentially turning the problem description into a solution. So at minimums you need some point for value of $\tilde S(t,x)$ for $t\in[0,1)$. The easiest construction is to give the pair $(x,t)$ as that value. Next you have to ensure continuity at $t=1$, which is satisfied by the equivalence relation. Then extend this to a proper map on the constructed space. – Lutz Lehmann Jun 08 '20 at 15:13
  • That's a clever approach and I now get the idea. But I still don't see how to prove the well-definedness of $\tilde{S}$ as I'm not sure how the equivalence relation looks like apart from the one relation you gave me ... – Rino Jun 08 '20 at 15:36
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    This is the only relation. All $x$ in $f^{-1}({y})$ give the equivalence $(x,1)\equiv (y,0)$. There should be no problem with $\tilde S$ being well-defined, as it only has images in $X\times [0,1)$ where no identification takes place. What is non-trivial is to show the continuity of it in all of its variables. – Lutz Lehmann Jun 10 '20 at 08:38
  • What metric is used on our new space $X \times [0,1]$ and what about the first proposed space $Y=C^0([0,1])$? Is there a way to extend $S$ to this space rather than $Y=X \times [0,1]$? @Lutz – Rino Jun 19 '20 at 09:15
  • The construction in this answer is known as the suspension operation in dynamical systems. Your hint to use $Y=C^0([0,1])$ seems off to me, in particular it violates the standard use of the $C^0$ notation, in which $C^0(A,B)$ denotes the set of continuous functions from a topological space $A$ to a topological space $B$. – Lee Mosher Jun 29 '20 at 15:31
  • @LeeMosher We often use $C^0([0,1])$ shortly for $C^0([0,1],\mathbb{R})$ or $C^0([0,1],X)$. The latter one is the only logical my professor probably meant. Thank you for the link, but I never heard of this concept. Do you know anything about the metric structure on Y? Probably the qoutient metric, right? – Rino Jun 29 '20 at 16:27