Let $X$ be open convex. There is a sequence $(Y_n)$ of bounded open convex sets such that $\overline{Y_n}\subset Y_{n+1}$ and $\bigcup_nY_n=X$ This thread is meant to record a question that I feel interesting during my self-study. I'm very happy to receive your suggestion and comments.


Theorem: Let $X  \subset \mathbb R^d$ be non-empty open convex. There is a sequence $(Y_n)$ of bounded open convex sets such that $\overline{Y_n} \subset Y_{n+1}$ and $\bigcup_n Y_n = X$.

 A: Let $d_{X^c} (x) = \min_{y \in X^c} \|x-y\|$. It is straightforward to show that $d_{X^c}$ is continuous and $d_{X^c}(x)>0$ iff $x \in X$ (since $X$ is open).
Define $Y_n = \{x \in X | d_{x \in X^c}(x) > {1 \over n} \}$. It is straightforward to show that $\overline{Y_n} = \{x \in X | d_{x \in X^c}(x) \ge {1 \over n} \} \subset Y_{n+1}$ and $\cup_n Y_n = X$.
To see that $Y_n$ is convex, suppose $x,y \in Y_n$ and $\delta = \min(d_{x \in X^c}(x), d_{x \in X^c}(y)) > {1 \over n}$. Then if $B$ is the open unit ball, we have
$\{x\} + \delta B \subset X$ and $\{y\} + \delta B \subset X$ and so $\lambda (\{x\} + \delta B) + (1-\lambda) (\{x\} + \delta B) = \{\lambda x + (1-\lambda)y\} + \delta B \subset X$. In particular, $d_{x \in X^c} (\lambda x + (1-\lambda)y\}) \ge \delta > {1 \over n}$.
A: Let
$$
X_n := \left \{x \in \mathbb R^d \,\middle\vert\, \inf_{y \in X^c} |x-y| > \frac{1}{n} \right \} \quad \forall n \in \mathbb N^*.
$$
Clearly, $X_n \subset X_{n+1} \subset X$.

*

*$X_n$ is open.

It suffices to prove that
$$
X'_n := \left \{x \in \mathbb R^d \,\middle\vert\, \inf_{y \in X^c} |x-y| \le \frac{1}{n} \right \} \quad \forall n \in \mathbb N^*.
$$
is closed. Let $(x_m) \subset X'_n$ such that $x_m \to a \in \mathbb R^d$. For all $n \in \mathbb N^*$, we have
$$
\begin{align}
\inf_{y \in X^c} |a-y| &\le \inf_{y \in X^c} [|x_m-y|+|x_m-a|] \\
&= \left [ \inf_{y \in X^c} |x_m-y| \right ] + |x_m-a| \\
&\le \frac{1}{n} + |x_m-a|.
\end{align}
$$
It follows that
$$
\inf_{y \in X^c} |a-y| \le  \frac{1}{n} + \lim_m|x_m-a| = \frac{1}{n}.
$$
Hence $a \in X'_n$.


*$\overline{X_n} \subset X_{n+1}$.

Let $(x_m) \subset X_n$ such that $x_m \to a \in \mathbb R^d$. For all $n \in \mathbb N^*$, we have
$$
\begin{align}
\inf_{y \in X^c} |a-y| &\ge \inf_{y \in X^c} [|x_m-y|-|x_m-a|] \\
&= \left [ \inf_{y \in X^c} |x_m-y| \right ] - |x_m-a| \\
&> \frac{1}{n} - |x_m-a|.
\end{align}
$$
It follows that
$$
\inf_{y \in X^c} |a-y| \ge  \frac{1}{n} - \lim_m|x_m-a| = \frac{1}{n} > \frac{1}{n+1}.
$$
It follows that $a \in X_{n+1}$.


*$\bigcup_n X_n = X$.

If $a \in X$, then $\inf_{y \in X^c} |a-y| >0$ because $X^c$ is closed. Then $a \in X_n$ for sufficiently large $n$.


*$X_n$ is convex.


Lemma: $a \in X_n \iff \overline B(a, 1/n) \subset X$.

Let $a,b \in X_n$. Then $\overline B(a, 1/n) \subset X$ and $\overline B(b, 1/n) \subset X$. Let $\lambda \in (0, 1)$ and $c := \lambda a + (1-\lambda)b$. Let $x \in \overline B(0, 1/n)$. By our Lemma, it suffices to prove $|x+c| \le 1/n$. Indeed,
$$
\begin{align}
|x+c| &= |x+ [\lambda a + (1-\lambda)b]| \\
&= |\lambda(x+a) + (1-\lambda)(x+b)| \\
&\le \lambda |x+a|+(1-\lambda) |x+b| \\
&\le \lambda \frac{1}{n} + (1-\lambda) \frac{1}{n} \\
&= \frac{1}{n}.
\end{align}
$$
Notice that $B(0, n)$ is open convex. Then $Y_n := X_n \cap B(0, n)$. Then $Y_n$ is bounded open convex and $\overline {Y_n} \subset Y_{n+1}$. Obviously, $\bigcup_n Y_n = X$. This completes the proof.
