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According to Theorems 1 and 3 in this review article we have

Weierstrass: Suppose $f$ is a continuous function on a closed bounded interval $[a,b] \subset\mathbb{R}$. For every $\epsilon > 0$ there exists a polynomial $p$ such that for all $x \in [a,b]$ we have $| f(x)− p(x)| < \epsilon$.

Mergelyan: If $K$ is a compact set in $C$ with connected complement, then every continuous function $f\colon K\to \mathbb{C}$ that is holomorphic in the interior of $K$ can be approximated uniformly on $K$ by holomorphic polynomials.

Both Wikipedia and the review say that the latter is a generalization of the former. In which sense is this true? How does Weierstrass' theorem follow from Mergelyan's?

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The hypotheses of Weierstrass satisfy the hypotheses of Mergelyan:

$K=[a,b] \subseteq \mathbb C$ is a compact set with connected complement.

Since $K$ has empty interior, "holomorphic in the interior of $K$" is true.

The conclusions of Mergelyan imply the conclusions of Weierstrass:

"approximated uniformly" is the same as "$| f(x)− p(x)| < \epsilon$ for all $x \in K$".

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    $\begingroup$ The issue is that those approximation polynomials are complex. To complete the argument, you might consider their real part. $\endgroup$ – mathcounterexamples.net Oct 5 '18 at 12:57
  • $\begingroup$ @mathcounterexamples.net, ah, I see, that's an important point, thanks. $\endgroup$ – lhf Oct 5 '18 at 12:58
  • $\begingroup$ Oh how embarassing. I somehow thought the interior of $K$ needed to be nonempty and thought that somehow the function from the Weierstrass theorem would need to be thought of as a function on a part of the boundary of a unit disk etc. $\endgroup$ – Bananach Oct 5 '18 at 13:24

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