$X$ be a compact Hausdorff space such that $\dim$ $C(X,\mathbb{R})<\infty$ we need to show $|X|<\infty$, I must say I have no idea how to prove this result. please help. Thank you!
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Can you show that you can always extend a function defined on a finite set of points on $X$ to a continuous function on all of $X$? – Zhen Lin Feb 07 '13 at 11:51
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1See also: http://math.stackexchange.com/q/250325/49437 for a generalization. – Martin Feb 07 '13 at 15:04
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Let $N$ be the dimension of $C(X,\Bbb R)$. Assume that there are $N+1$ distinct points $x_1,\dots,x_{N+1}$ in $X$. As $X$ is compact and Hausdorff it is normal, hence by the Urysohn lemma, we can find for each $j$ a continuous function $g_j\colon X\to\Bbb R$ such that $g_j(x_k)=0$ when $k\neq j$ and $g_j(x_j)=1$.
The family $\{g_j,1\leqslant j\leqslant N+1\}$ is necessarily linearly dependent, so we can assume that $g_{N+1}=\sum_{j=1}^Ng_j$. Evaluating this equality at $x_{N+1}$ yields a contradiction.
This proves that $\dim C(X,\Bbb R)=\operatorname{card}(X)$.
The result is not necessarily true when $X$ is not assumed to be Hausdorff. For example, take $X$ an infinite set with the topology $\{\emptyset,X\}$. The only continuous functions are constant ones.
Davide Giraudo
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2Perhaps it's worth pointing out where we used the compactness assumption... – Zhen Lin Feb 07 '13 at 13:12
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You are right. This is indeed where compactness is used (I should have done this edit one year ago...). – Davide Giraudo May 30 '14 at 09:51