Function mapping finite sets

[SlipEternal]

Let \(\displaystyle \displaystyle F:=\left\{f:[0,\infty)\to \mathbb{R}\mid \begin{array}{l}f(0) = \emptyset, \lim_{x\to \infty}{f(x)} = \mathbb{R}, \mbox{ and}\\ \forall x,y\in [0,\infty), f(x) \subset f(y), f(x) \neq f(y)\end{array}\right\}\). Does there exist \(\displaystyle f\in F\) such that for some \(\displaystyle x\in[0,\infty)\), \(\displaystyle f(x)\) is a nonempty, finite set?

 

[pka]

Let \(\displaystyle \displaystyle F:=\left\{f:[0,\infty)\to \mathbb{R}\mid \begin{array}{l}f(0) = \emptyset, \lim_{x\to \infty}{f(x)} = \mathbb{R}, \mbox{ and}\\ \forall x,y\in [0,\infty), f(x) \subset f(y), f(x) \neq f(y)\end{array}\right\}\). Does there exist \(\displaystyle f\in F\) such that for some \(\displaystyle x\in[0,\infty)\), \(\displaystyle f(x)\) is a nonempty, finite set?

I have reviewed hundreds of articles on these sorts of topics. But I must say that I have no idea what you are asking.
\(\displaystyle \mathcal{F}\) seems to be a collection of functions.
However, these functions seem to map numbers to subsets of \(\displaystyle \mathbb{R}\). BUT that is not clear in the OP.
Please review what you have posted. There seems to be some internal contradictions.

 

[SlipEternal]

I have reviewed hundreds of articles on these sorts of topics. But I must say that I have no idea what you are asking.
\(\displaystyle \mathcal{F}\) seems to be a collection of functions.
However, these functions seem to map numbers to subsets of \(\displaystyle \mathbb{R}\). BUT that is not clear in the OP.
Please review what you have posted. There seems to be some internal contradictions.

Oops, I meant:

\(\displaystyle \displaystyle \mathcal{F}:=\left\{f:[0,\infty)\to \mathcal{P}(\mathbb{R})\mid\begin{array}{l}f(0)=\emptyset, \lim_{x\to\infty}{f(x)}=\mathbb{R}\\ \forall x,y\in[0,\infty), f(x)\subset f(y) \mbox{ and }f(x)\ne f(y)\end{array}\right\}\)

So, does there exist \(\displaystyle f\in \mathcal{F}\) such that for some \(\displaystyle x\in[0,\infty)\), \(\displaystyle f(x)\) is nonempty, but finite?

 

[pka]

Now it seems that \(\displaystyle \mathcal{F}\) is the collection of all functions \(\displaystyle f:[0,\infty)\to\mathcal{P}(\mathbb{R})\), with \(\displaystyle f(\emptyset)=0\). In that case the answer is of course yes.

On the other hand, if \(\displaystyle \mathcal{F}\) is just some collection then no it does no have to contain such a function.

 

[SlipEternal]

Here is my reasoning for being skeptical:

Assume there does exist such a function. Let \(\displaystyle f:[0,\infty)\to\mathbb{R}\) be defined such that \(\displaystyle \displaystyle f(0)=\emptyset, \lim_{x\to\infty}{f(x)}=\mathbb{R}, \forall x,y\in [0,\infty), x<y \Rightarrow f(x) \subsetneq f(y)\). Assume there exists some \(\displaystyle a \in [0,\infty)\) such that \(\displaystyle f(a)\) is nonempty and finite. Consider, the sequence \(\displaystyle a_n=\frac{a}{n+1}\) (which approaches zero as \(\displaystyle n\) approaches infinity). \(\displaystyle f(a_n)\) must be an infinite sequence of properly nested sets (the cardinality of \(\displaystyle f(a_n)\) must be strictly larger than the cardinality of \(\displaystyle f(a_{n+1})\)). This seems a contradiction to the notion that there exists a value in \(\displaystyle [0,\infty)\) satisfying the property that the image is finite and nonempty.

 

[pka]

Here is my reasoning for being skeptical:
Assume there does exist such a function. Let \(\displaystyle f:[0,\infty)\to\mathbb{R}\) be defined such that \(\displaystyle \displaystyle f(0)=\emptyset, \lim_{x\to\infty}{f(x)}=\mathbb{R}, \forall x,y\in [0,\infty), x<y \Rightarrow f(x) \subsetneq f(y)\). Assume there exists some \(\displaystyle a \in [0,\infty)\) such that \(\displaystyle f(a)\) is nonempty and finite. Consider, the sequence \(\displaystyle a_n=\frac{a}{n+1}\) (which approaches zero as \(\displaystyle n\) approaches infinity). \(\displaystyle f(a_n)\) must be an infinite sequence of properly nested sets (the cardinality of \(\displaystyle f(a_n)\) must be strictly larger than the cardinality of \(\displaystyle f(a_{n+1})\)). This seems a contradiction to the notion that there exists a value in \(\displaystyle [0,\infty)\) satisfying the property that the image is finite and nonempty.

Now you have added a new condition.
\(\displaystyle x<y \Rightarrow f(x) \subsetneq f(y)\) that was not part of the OP.
Now the answer is no. A simple cardinality argument is enough.

 

[SlipEternal]

Oops, the way I wrote it in the OP, \(\displaystyle \mathcal{F}=\emptyset\). I meant \(\displaystyle \forall x,y\in [0,\infty)\mbox{ s.t. }x<y...\)