Distance to closed and convex sets that have intersection $C \subseteq D \rightarrow$ $ d_D(x^*) \leq d_C(x^*) $

Question

Let $C \subseteq \mathbb{R}^n$ be a closed convex set, and $x^* \in C^c$ (not in $C$ and its closure).

Define the Euclidean distance from $x^*$ to $C$ as $d_C(x^*):=\min_{z \in C}\|z -x^*\|_2$.

Let $D$ be a closed convex set containing $C$, i.e., $C \subseteq D$.

Show that $$ d_D(x^*) \leq d_C(x^*) $$

I do not know how to use $C \subseteq D$ together with taking minimum.

Answer 1

Since $C \subseteq D$, the minimum over $C$ can only be greater or equal to the minimum over $D$, so

$$d_D(x^*)=\min_{z \in D}\|z -x^*\|_2 \leq \min_{z \in C}\|z -x^*\|_2= d_D(x^*)$$

Answer 2

for $\forall z \in D$ and $y\in C$, we have \begin{align} d_D(x^*) &= min_{z \in D} \|z-x^*\|_2 \\ &= min_{y\in C} \|(z_{best} -y)+ (y-x^*)\|_2 \\ &\leq min_{y \in C}\|z_{best}-y\|_2 + min_{y \in C}\|y-x^*\|_2 \\ & \leq min_{y\in C}\|y-x^*\|_2 \\ &=d_C(x^*) \end{align}

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Answer above has mistakes. while, maybe we can try a new way to solve this.

Notice that for any element $y \in C$, because $C \subseteq D$, so that $y \in D$, and another element $z_{best} \in D$, satisfy $d_D(x^*)=min_{z \in D} \|z-x^*\|_2=\|z_{best}-x^*\|_2$, from this definition, it is quite clear to see $$ \|y-x^*\|_2 \geq \|z_{best}-x^*\|_2, \forall y \in C$$ which results in $$d_C(x^*) = min_{y\in C}\|y-x^*\|_2 \geq \|z_{best}-x^*\|_2 = d_D(x^*)$$