$C(X)$ and $C^{\ast}(X)$ for various subspace of $\mathbb{R}$

Question

The set $ C= C ( X ) $ of all continuous, real valued functions on a topological space$ X$ will be provided with an algebraic structure and an order structure.

Addition and multiplication are defined by the formulas.

$ ( f + g ) = f( x) + g( x) $, and$ (fg) (x) = f(x)g(x) $

define:

$C^{*} = C^{*}( X ) = \{ f \in C ( X ) \vert \quad f\quad is \quad bounded \}$

consider the subspace $\mathbb{R}$ , $ \mathbb{Q}$, $\mathbb{N}$ and $ \mathbb{N}^{*} = \{ 1, 1/2, .........., 1/n ,...... 0 \} $ of $ \mathbb{R}$ , and the ring $ C $ and $ C ^{*}$ for each of these spaces. each of these rings is of cardinal $ c$.

my questions are:

1: for each $^{ m \leq \mathbf{N}_{0}}$ each ring on $\mathbb{R}$ , $\mathbb{N}$ or $ \mathbb{N}^{*} $ contains having exactly $ 2^{m}$ square rootsl . if a member of $ C ( \mathbb{Q} )$ has more than one square root, it has $ c $ of them.

2:each of $ C ( \mathbb{Q} )$, $ C ( \mathbb{N} )$ is isomorphic with a direct sum of two copies of itself. $ C ( \mathbb{N}^{*} )$ is isomorphic with a direct sum of two copies of two subrings, just one of which is isomorphic with $ C ( \mathbb{N}^{*} )$.

3: the ring $ C ( \mathbb{R})$ is isomorphic with a proper subring.

Answer 1

Ad 1

Consider a function $h:[0,1]\to\mathbb{R}$ given by $h(x)=x-x^2$. I'm looking for a function such that $h(0)=h(1)=0$ and $h(x)>0$ otherwise. Note that there are two continuous solutions for $f^2=h$ namely $f_1(x)=\sqrt{h(x)}$ and $f_2(x)=-\sqrt{h(x)}$.

Now for $n\in\mathbb{N}$ define

$$G_n:\mathbb{R}\to\mathbb{R}$$ $$G_n(x)=\begin{cases} h\big(x-\lfloor x\rfloor\big) &\mbox{if } 0\leq x\leq n \\ 0& \mbox{otherwise} \end{cases}$$

where $\lfloor\cdot\rfloor$ is the floor function. So I'm shifting and gluing copies of $h$ $n$-times. It is continuous and bounded.

You can easily check that the equation $f^2=G_n$ has exactly $2^n$ solutions. Each solution is given by gluing together $n$-times shifted $f_1$ and/or $f_2$. Thus every solution corresponds to a function $s:\{0,1,\ldots,n\}\to\{f_1, f_2\}$ which corresponds to a subset of $\{0,1,\ldots,n\}$.

Note that $G_{\infty}(x)=h\big(x-\lfloor x\rfloor\big)$ has exactly $2^{\aleph_0}$ solutions because again every solution corresponds to a subset of $\mathbb{Z}$.

The same works for $C(\mathbb{N})$ and something analogous can be constructed for $C(\mathbb{N}^*)$ by "shrinking" gluing intervals and scaling down values at each (so that the limit at $0$ is $0$).

Case $C(\mathbb{Q})$: Let $f_1, f_2$ be two square roots of $g\in C(\mathbb{Q})$. Note that the set $\{x\ |\ f_1(x)\neq f_2(x)\}$ is nonempty and open, i.e. it contains an interval, say $I=(a,b)\cap\mathbb{Q}$. Now for any $r\in(a,b)$, $r\in\mathbb{R}\backslash\mathbb{Q}$ (note that $r$ is not rational) define

$$f_r(x)=\begin{cases} f_1(x) &\mbox{if } x<r \\ f_2(x) &\mbox{otherwise} \\ \end{cases} $$ Obviously $f_r^2=g$ and $f_r\neq f_s$ for $r\neq s$. Since $r$ is not rational then $f_r$ is additionally continuous. Obviously it is bounded if $g$ is.

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Ad 2 This follows from the fact that both $\mathbb{Q}$ and $\mathbb{N}$ can be divided into two topological copies (i.e. can be expressed as a disjoint union of subsets):

• Pick a non-rational $r\in\mathbb{R}$. Then $\mathbb{Q}=\big((-\infty,r)\cap \mathbb{Q}\big) \cup \big((r,\infty)\cap \mathbb{Q}\big)$. This decomposition allows us to look at every continous map $f:\mathbb{Q}\to\mathbb{R}$ as a pair of continous maps.
• Analogously $\mathbb{N}=2\mathbb{N}\cup(\mathbb{N}\backslash 2\mathbb{N})$.

Note that the same argument doesn't work for $\mathbb{R}$ because such pair of functions does not have to produce continuous map when combined.

For $\mathbb{N}^*$ take $\mathbb{N}^*=\big(\mathbb{N}^*\backslash \{1\}\big)\cup\{1\}$. Again every function $\mathbb{N}^*\to\mathbb{R}$ can be easily expressed as a pair of function $\mathbb{N}^*\backslash\{1\}\to\mathbb{R}$ and $\{1\}\to\mathbb{R}$ and $C(\{1\})$ is not isomorphic to $C(\mathbb{N}^*)$.

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Ad 3 TODO: I'm still not sure about this one. Maybe someone else can fill the gap. Or perhaps you should ask it as a separate question. Maybe each one of those points should be a separate question?

Answer 2

You seem to be working from Gillman and Jerrison, "rings of continuous functions", a classic in topology..

As to 1. if a function $f$ is defined on $m$ closed intervals on which it is $> 0$ (and $0$ on the boundaries) ,then it has $2^m$ many square roots (on every part we can pick the $+\sqrt{f}$ or $-\sqrt{f}$ part independently. For $\mathbb{Q}$ and $m=\aleph_0$ we can use $n\sqrt{2}$ ,$n \in \mathbb{Z}$ as boundaries and use constant functions inbetween. (no necessity of value $0$ on boundaries there).

Those function can be bounded or unbounded as desired. @freakish has a nice argument why $>1$ square root implies $\mathfrak{c}$ many on $\mathbb{Q}$.

As to 2. We can write as $\mathbb{N} = \mathbb{N_1} \cup \mathbb{N_2}$ (even and odd say) and the map $f \to (f|_\mathbb{N_1},f|_\mathbb{N_2})$ is a ring isomorphism between $C(\mathbb{N})$ and $C(\mathbb{N}) \oplus C(\mathbb{N})$.

Also $\mathbb{Q} = \mathbb{Q} \cap (-\infty, \sqrt{2}) \cup \mathbb{Q} \cap (\sqrt{2}, +\infty)$, which are two homeomorphic disjoint open copies of $\mathbb{Q}$ and gives rise to a similar isomorphism between $C(\mathbb{Q})$ and $C(\mathbb{Q}) \oplus C(\mathbb{Q})$.

We can write $\mathbb{N}^\ast$ as the disjoint union of $A=\{1\}$ and $B = \mathbb{N}^\ast \setminus \{0\}$, both of which are open and $B \simeq \mathbb{N}^\ast$, so that $C(\mathbb{N}^\ast) \simeq C(\mathbb{N}^\ast) \oplus \mathbb{R}$, and also for bounded functions.

We cannot write $C(\mathbb{R})$ as such a disjoint sum of rings by connectedness.

As to 3. $\{f \in C(\mathbb{R}): f|_{[0,1]} \text{is constant} \}$ is a subring of $C(\mathbb{R})$ that is isomorphic to the whole ring, as we just split up a function on $0$: The isomorphism $T$ is given by $f \to g=T(f)$ where $g(x) = f(x) \text{ for } x < 0$, $g(x) = f(0) \text{ for } x \in [0,1]$, $g(x) = f(x-1) \text{for } x > 1$, which is continuous (i.e. $g$ is when $f$ is) and preserves algebraic operations (i.e. the map $f \to g$).