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For $\lambda \notin K$, write $g_{\lambda }(z)=(z-\lambda {)}^{-1}$. If $\lambda \in \Lambda$ andthenuniformly on $K$. Uniformly polynomially approximable functions form an algebra closed under uniform limits, so $\mu \in \Lambda$. Hence $\Lambda$ is open.

The set $\Lambda$ is also closed relative to $\mathbb{C}\setminus K$. Indeed, if ${\lambda }_j\in \Lambda$ and ${\lambda }_j\to \lambda \notin K$, thenso $g_{\lambda }$ is again uniformly approximable by polynomials. Since $K$ is closed, every point of $\Gamma$ has a neighbourhood in $\mathbb{C}\setminus K$; relative closedness of $\Lambda$ then gives a smaller neighbourhood disjoint from $\Lambda$. Thus both $\Lambda$ and $\Gamma$ are open.

It is not always true that $\Lambda$ is nonempty. Take $K=\mathbb{R}$. Any polynomial uniformly close on $\mathbb{R}$ to a bounded resolvent $g_{\lambda }$, with $\lambda \notin \mathbb{R}$, must itself be bounded on $\mathbb{R}$, and hence must be constant. Uniform limits of constants are constant, whereas $g_{\lambda }$ is not. Therefore

Boundedness of $K$ does not force $\Gamma$ to be empty either. Take $K=\{z:|z|=1\}$ and $\lambda =0$. If polynomials $p_n$ converged uniformly to $1/z$ on $K$, then uniform convergence and contour integration would givea contradiction. Hence

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.