Comparison between $\sup W$ and $\sup \overline{W}$

Question

It is well-known that $\sup S= \sup \overline{S}$ for any nonempty $S\subset \mathbb{R}$.

Now, let $W$ be a nonempty subset of $\mathbb{C}^d$ and $\overline{W}$ be the closure of $W$ with respect to the topology of $\mathbb{C}^d$.

Is $$\sup\{\|\lambda\|_2,\;\lambda=(\lambda_1,\cdots,\lambda_d) \in W\}=\sup\{\|\lambda\|_2,\;\lambda=(\lambda_1,\cdots,\lambda_d) \in \overline{W}\}\;?,$$ where $$\|\lambda\|_2:=\left(\displaystyle\sum_{k=1}^d|\lambda_k|^2\right)^{1/2}.$$

Clearly $$\sup\{\|\lambda\|_2,\;\lambda=(\lambda_1,\cdots,\lambda_d) \in W\}\leq\sup\{\|\lambda\|_2,\;\lambda=(\lambda_1,\cdots,\lambda_d) \in \overline{W}\}\;.$$

Answer 1

The function $\mathbb{C}^d\to\mathbb{R}$, $\lambda\mapsto\|\lambda\|_2$ is continuous.

If $W$ is unbounded, then $\sup\{\|\lambda\|_2:\lambda\in W\}=\infty$ and the same is obviously true for $\overline{W}\supseteq W$.

Suppose $W$ is bounded and set $s=\sup\{\|\lambda\|_2:\lambda\in W\}$. Let $\lambda\in\overline{W}$; then there exists a sequence $(\lambda_n)$ in $W$ converging to $\lambda$; then, by continuity, $\|\lambda\|=\lim_{n\to\infty}\|\lambda_n\|$; since $\|\lambda_n\|\le s$ for all $n$, also $\|\lambda\|\le s$.

Hence $\sup\{\|\lambda\|_2:\lambda\in \overline{W}\}\le s$. The converse inequality follows from $W\subseteq\overline{W}$.

Answer 2

Yes, they are equal. Suppose otherwise. That is, suppose that$$\sup\left\{\lVert\lambda\rVert_2\,\middle|\,\lambda\in W\right\}<\sup\left\{\lVert\lambda\rVert_2\,\middle|\,\lambda\in\overline W\right\}.$$Let$$s=\sup\left\{\lVert\lambda\rVert_2\,\middle|\,\lambda\in W\right\}\text{ and let }S=\sup\left\{\lVert\lambda\rVert_2\,\middle|\,\lambda\in\overline W\right\}.$$You are assuming that $s<S$. There is some $\lambda\in\overline W$ such that$$S-\lVert\lambda\rVert_2<\frac{S-s}2\tag1$$and there is some $\lambda^\ast\in W$ such that $\lVert\lambda-\lambda^\ast\rVert_2<\frac{S-2}2$, which implies that$$\lVert\lambda\rVert_2-\lVert\lambda^\ast\rVert_2<\frac{S-s}2.\tag2$$But then it follows from $(1)$ and $(2)$ that $S-\lVert\lambda^\ast\rVert_2<S-s$ and therefore $\lVert\lambda^\ast\rVert_2>s$, which is impossible by the definition of $s$ and because $\lambda^\ast\in W$.