To prove that the closure of a convex set is itself convex, we proceed as follows:

Definitions and Setup:

1. Let $C \subseteq \mathbb{R}^n$ be a convex set.

   • $C$ is convex if for all $x, y \in C$ and for all $\lambda \in [0,1]$: $\lambda x + (1-\lambda)y \in C.$

2. The closure of $C$, denoted $\overline{C}$, is the set of all limit points of $C$. Formally: $\overline{C} = C \cup \{\text{all points that are limits of sequences in } C\}.$

We aim to prove that $\overline{C}$ is also convex.

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Proof:

Step 1: Take arbitrary points in the closure of $C$:

Let $x, y \in \overline{C}$. By definition of the closure, there exist sequences $\{x_k\}$ and $\{y_k\}$ in $C$ such that: $x_k \to x \quad \text{and} \quad y_k \to y \quad \text{as } k \to \infty.$

Step 2: Consider a convex combination in $\overline{C}$:

For $\lambda \in [0,1]$, we need to show that the convex combination $z = \lambda x + (1-\lambda)y \in \overline{C}$.

Step 3: Use the sequences to construct the convex combination:

Define a new sequence $\{z_k\}$ in $C$ as: $z_k = \lambda x_k + (1-\lambda)y_k.$ Since $x_k, y_k \in C$ and $C$ is convex, $z_k \in C$ for all $k$.

Step 4: Show that the sequence converges to $z$:

$z_k = \lambda x_k + (1-\lambda)y_k \to \lambda x + (1-\lambda)y = z \quad \text{as } k \to \infty,$ because the operations of addition and scalar multiplication are continuous.

Step 5: Conclude $z \in \overline{C}$:

By definition of the closure, if a sequence in $C$ converges to $z$, then $z \in \overline{C}$.

Thus, $\lambda x + (1-\lambda)y \in \overline{C}$.

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Final Conclusion:

Since $x, y \in \overline{C}$ were arbitrary and $\lambda x + (1-\lambda)y \in \overline{C}$, we conclude that $\overline{C}$ is convex.

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Recap:

• $\overline{C}$ contains all limit points of $C$.
• Convex combinations of points in $\overline{C}$ remain in $\overline{C}$, due to the continuity of sequences and convexity of $C$.

This proves that the closure of a convex set is convex.

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Do you want further clarification or details on any part of this proof? Here are some related questions:

1. Why is continuity of addition and scalar multiplication essential in this proof?
2. Can you prove the closure of a set is unique for a given set?
3. What happens if $C$ is not convex? Is $\overline{C}$ necessarily convex?
4. How does the topology of $\mathbb{R}^n$ influence the closure's properties?
5. How does this proof generalize to infinite-dimensional spaces like Hilbert spaces?

Tip: Visualizing convex sets as regions where line segments between any two points lie entirely inside the set can make the proof more intuitive!