Is the set $\left\{ \boldsymbol{\beta} \in \mathbb{R}^{p}: ||\mathbf{A} \boldsymbol{\beta} - \boldsymbol{d}|| \leq D \right\}$ compact? As the title of the question suggests, I am hoping to establish whether the set
$$\left\{ \boldsymbol{\beta} \in \mathbb{R}^{p}: ||\mathbf{A} \boldsymbol{\beta} - \boldsymbol{d}|| \leq D \right\}$$
is compact. $\mathbf{A} \in \mathbb{R}^{k+p}$ is a matrix of full column rank and $\mathbf{d} \in \mathbb{R}^{k}$ is a vector. I believe that since this is a bounded set (by the full column rank hypothesis) and the function $\boldsymbol{\beta} \mapsto || \mathbf{A} \boldsymbol{\beta} - \boldsymbol{d}||$ is continuous this is enough to establish compactness. Could you please tell me if my reasoning is correct?
Thank you.

You are correct. The set is bounded, as you say because $\mathbf A$ has full column rank (*) and closed because it is the preimage of the closed set $[0, D]$ by a continuous map.
(*) Why does this imply that the set is bounded? Let $\mathbf B \in \mathbb R^{p \times k}$ be such that $\mathbf B \mathbf A = I$ Then $$\Vert \beta \Vert = \Vert \mathbf B \mathbf A \beta \Vert\leq \Vert \mathbf B \Vert \cdot \Vert \mathbf A \beta\Vert$$
so that $\beta$ remains bounded when $\mathbf A \beta$ is bounded.
