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Let X be a compact metric space and ##C(X)## be the set of all continuous functions ##X \to \mathbb{R}##, equipped with the uniform norm, i.e. the norm defined by ##\Vert f \Vert = \sup_{x \in X} |f(x)|##

Note that this is well defined by compactness. Then, for a subset ##A \subset C(X)##, we define the uniform closure as the set ##\overline{A}## with respect to the uniform norm.

Now, in a proof I'm going through, it is claimed that if ##A## is an algebra (subvectorspace + closed under pointwise multiplication), then ##\overline{A}## is an algebra too.

I decided to prove this, and did the following:

Let ##f,g \in \overline{A}##. There are sequences of functions such that ##f_n \to f, g_n \to g## for the uniform norm. But convergence for the uniform norm is the same thing as uniform convergence of sequences of functions.

So, let ##\epsilon > 0##. Choose ##n_0## such that for all ##n \geq n_0, x \in X##, we have both

##|f_n(x) - f(x)| < \epsilon## and ##|g_n(x) - g(x)| < \epsilon##

Then, for ##n \geq n_0, x \in X##, we have:

##|f_n(x)g_n(x) - f(x)g(x)| \leq |f_n(x)||g_n(x)-g(x)| + |f_n(x) -f(x)||g(x)| \leq \Vert f_n \Vert \epsilon + \Vert g \Vert\epsilon < M \epsilon## for some number M (because all the functions are bounded, as they are continuous.

This shows that ##f_ng_n \to fg##, and hence ##fg\in \overline{A}## as ##f_ng_n \in A## because it is an algebra.

Analoguous, we can prove that linear combinations remain in the set. Is this a correct proof?

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# I If A is an algebra, then its uniform closure is an algebra.

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