enter image description here enter image description here> Theorem: Let $A \subset C(K)$ such that

  1. $A$ is a subalgebra with unity $1$

  2. For each $x, y \in K $ with $x \neq y $, there exists $f \in A$ such that $f(x)\neq f(y)$.

Then $ \overline A = C(K)$, where $C(K)$ is the space of continuous functions over the compact space $K$.

Proof: We have to show that for every function $f\in C(K)$, for every $x\in K$ and for every $\epsilon>0$, there exists $g_x \in \overline A$ such that $g_x\le f+\epsilon$ and $g(x)>f(x)$.

My question is: why do we want to show that? Why is the above statement equivalent to the theorem?

  • $\begingroup$ In the "proof" part, shouldn't it be "$g_x(x)>f(x)$"? $\endgroup$ – N.Bach Apr 25 '17 at 17:21
  • $\begingroup$ @N.Bach nop, in the book was g(x)>f(x) $\endgroup$ – Aaron Martinez Apr 25 '17 at 18:10
  • $\begingroup$ @N.Bach actually it wasn't a book it was a pdf file, so maybe it's wrong and you're correct $\endgroup$ – Aaron Martinez Apr 25 '17 at 18:11
  • $\begingroup$ @N.Bach why do you think it should be $g_x$(x)>f(x) ? $\endgroup$ – Aaron Martinez Apr 25 '17 at 18:12
  • 1
    $\begingroup$ the function $g(x):=P(f(x))$ where $P$ is a polynomial, hence $g$ have the form $$g(x)=\sum_{k=0}^n a_k f(x)^k$$ for some constants $a_k$. And because $\overline A$ is an algebra then each monomial $f^k$ belong to $\overline A$, and $g$ belong to $\overline A$ (check the definition of algebra). $\endgroup$ – Masacroso Apr 25 '17 at 20:04

A subset $A$ of a metric space $M$ is dense if and only if $\overline A=M$.

This imply, because $M$ is a metric space, that for every point $x\in M$ and every $\epsilon>0$ then the open ball $\Bbb B(x,\epsilon)$ contains points of $A$.

Hence there is a sequence on $A$ that converges to $x$, for any $x\in M$.

Now, the metric space of continuous functions $C(\Bbb K)$ is generally defined with the norm $\|{\cdot}\|_\infty$. Then if $A$ is dense in $C(\Bbb K)$ then for every $f\in C(\Bbb K)$ exists a sequence $(f_n)$ in $A$ that converges to $f$, that is, for any $\epsilon>0$ exists a $N\in\Bbb N$ such that

$$\|f_n-f\|_\infty<\epsilon,\quad\forall n\ge N\tag{1}$$

Of course is possible that the sequence $(f_n)$ approach to $f$ from above, that is when $f_n(x)>f(x)$ for all $x\in\Bbb K$. But without more explanation, the statement (and the notation)

$$g_x\le f+\epsilon\tag{2}$$ is not so clear. However if we substitute the space of continuous functions by the space of continuous bounded functions then the reversed triangle inequality


is meaningful because $\|f_n\|_\infty$ and $\|f\|_\infty$ are finite. Then if the sequence $(f_n)$ approaches from above to $f$ we can write

$$\|f_n\|_\infty<\epsilon+\|f\|_\infty$$ what is probably what means $(2)$.

  • $\begingroup$ Do you think it should be $g_x$ instead of g? $\endgroup$ – Aaron Martinez Apr 25 '17 at 19:01
  • $\begingroup$ @Aaron and what is $g$ then? I cant say if it must be $g_x$ or $g$ because I need more context. It is not enough what you showed in the question to guess the meaning of the notation. $\endgroup$ – Masacroso Apr 25 '17 at 19:16
  • $\begingroup$ ok, I'll post a photo of the whole proof then $\endgroup$ – Aaron Martinez Apr 25 '17 at 19:18
  • $\begingroup$ and now how you explain me that "We have to show that for every function $f\in C(K)$, for every $x\in K$ and for every $\epsilon>0$, there exists $g_x \in \overline A$ such that $g_x\le f+\epsilon$ and $g(x)>f(x)$." is equivalent to A is dense in C(K) ? $\endgroup$ – Aaron Martinez Apr 25 '17 at 23:50

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