# What if every function in $C(X)$ has finite spectrum?

Suppose that $X$ is a compact Hausdorf space, and that every continuous function on $X$ has finite range. How do I conclude that $X$ is a finite set, hence with discrete topology? So far, I have managed to use Urysohn to show that the closed maximal connected component decomposition of $X$ (as opposed to the path component decomposition) is one where every component is a 1 point set.

This question arose from trying to solve the question that every C* algebra with all normal elements having finite spectra are finite dimensional. A more direct solution to this general question would also suffice, because actually implies the special case I've reduced it to. By taking a unitization, and then a MASA in the unitization, one gets to the case I've gotten to.

Also, a topological question that is in the same setup but handles a related operator algebras question is this: if a net $u_i$ and $u$ in $X$ has the property that for all continuous $f \in C(X)$ one has that $f(u_i)$ converges to $f(u)$ then must $u_i$ converge to $u$ in topology? This is for showing that the gelfand representation of a $C(X)$ space is itself. I'd like this fact for $X$ locally compact Hausdorf, actually.

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For 2: Suppose $u_i$ doesn't converge to $u$. Then there is a neighborhood $U$ of $u$ and a subnet such that $u_j \notin U$ for all $j$ (since $u_j$ is not eventually in $U$, it is frequently in $U^c$). Apply Urysohn's lemma to get a continuous function such that $f(u) = 1$, supported in $U$. Get $f(u_j) = 0$ for all $j$, so $0 = f(u_j) \nrightarrow f(u) = 1$. –  commenter Oct 3 '12 at 8:09

For the first question. suppose X has an infinite sequence of point $x_0, ...,x_n,...$.

for each $i$ considere a function $f$ such that $f_i(x_0)= 0$,...,$f_i(x_{i-1})=0$, $f_i(x_i)=1$, and $\Vert f_i \Vert \leqslant 1$

(Uryshon can do that, just consider the function on {x_0...x_i} which is a discret space and then use the extension properties...)

Let $g = \sum \frac{f_i}{2^i}$.

then $\frac{1}{2^{i-1}} \geqslant g(x_i)\geqslant \frac{1}{2^i}$

ensure you that $g$ takes an infinite number of value.

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Yes, because each f_i can be taken to be nonnegative. –  Jeff Oct 3 '12 at 7:56
Yes you are right, in my mind they were positive but i forgot to write it ^^ –  Simon Henry Oct 3 '12 at 10:57

For your first question, let $x\in X$ be arbitrary, and let $U$ be any open set containing $x$. Let $f:X\to[0,1]$ be a continuous function such that $f(x)=0$ and $f(y)=1$ for all $y\in X\setminus U$, and let $\alpha$ be the smallest non-zero element of the range of $f$. Then $V=\left\{y\in X:f(y)<\frac{\alpha}2\right\}$ is a clopen neighborhood of $x$ contained in $U$. Thus, $X$ has a clopen base, i.e., is zero-dimensional. This means that if $p$ is a non-isolated point of $X$, we can construct a sequence $\langle U_n:n\in\Bbb N\rangle$ of clopen neighborhoods of $p$ such that $U_0=X$, $U_{n+1}\subseteq U_n$ for each $n\in\Bbb N$, and for each $n\in\Bbb N$ there are distinct points $x_n,y_n\in U_n\setminus U_{n+1}$. Let $K=\bigcap_{n\in\Bbb N}U_n$.

For each $n\in\Bbb N$ there is a continuous function $$f_n:U_n\setminus U_{n+1}\to\left[ \frac1{2^{n+1}},\frac1{2^n}\right]$$ such that $f(x_n)=\dfrac1{2^n}$ and $f(y_n)=\dfrac1{2^{n+1}}$. Define

$$f:X\to[0,1]:x\mapsto\begin{cases} f_n(x),&\text{if }x\in U_n\setminus U_{n+1}\\ 0,&\text{if }x\in K\;. \end{cases}$$

(In other words, $f=\sum_{n\in N}f_n$ on $X\setminus K$, and $f[K]=\{0\}$.) Clearly $f$ is continuous on $U\setminus K$, since each part of its definition has clopen domain, and the construction also ensures that $f$ is continuous on $K$. But the range of $f$ includes $2^{-n}$ for every $n\in\Bbb N$, so it’s infinite. Thus, $X$ cannot have a non-isolated point and must therefore be finite.

Added: For the second question, suppose that the net $\nu=\langle x_i:i\in I\rangle\not\to x$, where $\langle I,\le\rangle$ is a directed set. Then $x$ has an open neighborhood $U$ such that $\nu$ is frequently in $X\setminus U$. Let $f:X\to[0,1]$ be a continuous function such that $f(x)=0$ and $f(y)=1$ for $y\in X\setminus U$, and suppose that $\langle f(x_i):i\in I\rangle\to 0=f(x)$. Then there is an $i_0\in I$ such that $f(x_i)<\frac12$ for all $i\ge i_0$. Clearly $\left\{y\in X:f(y)<\frac12\right\}\subseteq U$, so there is an $i\ge i_0$ such that $x_i\notin U$ and hence $f(x_i)=1$; this contradiction shows that $\langle f(x_i):i\in I\rangle\not\to f(x)$.

In other words, if $\langle f(x_i):i\in I\rangle\to f(x)$ for every $f\in C(X,\Bbb R)$, then $\langle x_i:i\in I\rangle\to x$ in $X$.

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For your second question, if X is compact, you can extract a converging subnet from (u_i) and let u' be its limit... you have f(u)=f(u') for every continuous f, and hence u=u'. If X is just locally compact, then consider its one point 'alexandrov' compactification. (and if he is just regular, then the Stone Cech compactification will do the trick).

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I think this only shows that $u=u'$, not that this common value is the limit of the net. Also, I should note that when I said I wanted this for X locally compact Hausdorf, let me clarify in case there is any misunderstanding that the relevant object then is not $C(X)$ but rather $C_0(X)$ the continuous functions vanishing at infinity. –  Jeff Oct 3 '12 at 7:55
Yes, but if $(u_i)$ was not converging to $u$, then you can construct a sub net of $u_i$ which stay away from $u$, and hence extract a "sub-sub-net" which converge to something diferent than u, which leads to a contradiction. that a classical fact in topologie : in a compact set, if every convergent extration of a sequence/net converge to the same value, then the sequence/net is convergente. –  Simon Henry Oct 3 '12 at 10:55
and the argument still hold for $C_0(X)$ on a localy compact space as $C_0(X)$ separate the point of the alexandrov compactification of $X$. ($C_0(X)$ is the set of function on the compactification which vanish at infinity... ) –  Simon Henry Oct 3 '12 at 10:58
I have a problem with this fact. In a metric space, one has a convergent sequence whenever he verifies that every subsequence has a convergent subsequence converging to the same value. But with nets, I don't see an analogous result, because for me a subnet is the original net precomposed with an increasing cofinal function. If something like this still goes through, then the construction must be quite strange. I am quite curious about the details of this, but at least for this problem the suggested answer in the comment also works if you just use urysohn w/o subnets. –  Jeff Oct 3 '12 at 15:13
Well, if you have a net $(u_i)$ which does not converge to a value $x$. Then by (the negation of the) definition there exist a neighborhood $V$ of $u$ such that for every $i \in I$ there exist $j \in J$ such that $u_j$ is not in $V$. Hence the set of $i$ such that $u_i$ is not in $V$ is cofininal in $I$ and it gives us a sub net (the increasing function you're talking about is just the inclusion) of $u_i$ such that no "sub-sub-net" converge to $x$. which leads to a contradiction if the space is compact and if every converging subnet of $(u_i)$ converge to $u$. –  Simon Henry Oct 4 '12 at 8:24