Closure of a cone

Question

Suppose a set $\mathcal{X}$ is closed and bounded, and define $\mathcal{K}_\mathcal{X} = \{(x,t) : t > 0, \frac{x}{t} \in \mathcal{X} \}$. Show that:

$$\bar{\mathcal{K}}_\mathcal{X} = \mathcal{K}_\mathcal{X}\cup \{(0,0) \}$$

My attempt at Solution :

For any $\mathcal{X}$, $\mathcal{K}_\mathcal{X}$ is a cone. The closure of any cone must contain the origin. So this implies $\mathcal{K}_\mathcal{X} + \{(0,0) \} \subseteq \bar{\mathcal{K}}_\mathcal{X}$.

Now, I want to show $\mathcal{K}_\mathcal{X} + \{(0,0) \} \supseteq \bar{\mathcal{K}}_\mathcal{X}$.

Since $\mathcal{X}$ is bounded and closed, there exists an $M > 0$ such that $\forall$ $x \in \mathcal{X}$, we have $||x|| \leq M$. This lets us write $\mathcal{K}_\mathcal{X}$ as:

$\mathcal{K}_\mathcal{X} = \{(x,t) : t > 0, t \geq \frac{||x||}{M}, \frac{x}{t} \in \mathcal{X} \}$.

Case 1: $\mathcal{X}$ does not contain the origin 0. So, we can write

$\mathcal{K}_\mathcal{X} = \{(x,t) : t > 0, t \geq \frac{||x||}{M}, \frac{x}{t} \in \mathcal{X} \}$ as simply:

$\mathcal{K}_\mathcal{X} = \{(x,t) :, t \geq \frac{||x||}{M}, \frac{x}{t} \in \mathcal{X} \}$ which is closed.

So $\bar{\mathcal{K}_\mathcal{X}} = \mathcal{K}_\mathcal{X} \subseteq \mathcal{K}_\mathcal{X} \cup \{(0, 0) \}$

Case 2: $\mathcal{X}$ contains the origin $0$. let $\mathcal{X}' = \mathcal{X} \backslash \{0 \}$.

Then, we can write $\mathcal{K}_\mathcal{X} = \{(x,t) : t \geq \frac{||x||}{M}, \frac{x}{t} \in \mathcal{X}' \} \cup \{(0,t) : t > 0 \} = S_1 \cup S_2$. (Equation 1)

The closure of a union of two sets is equal to the union of the closures. So, we have:

$\bar{\mathcal{K}}_\mathcal{X} = \bar{S_1} \cup \bar{S_2}$. (Equation 2)

Now, $\bar{S_2} = \{(0,t) : t \geq 0 \} $.

But $S_1$ could be open or closed. If it is closed, then $\bar{S_1} = S_1$. in which case we have:

$\bar{\mathcal{K}}_\mathcal{X} = \{(x,t) : t \geq \frac{||x||}{M}, \frac{x}{t} \in \mathcal{X}' \} \cup \{(0,t) : t \geq 0 \}$. By comparing with (Equation 1), this is just saying:

$\bar{\mathcal{K}}_\mathcal{X} = \mathcal{K}_\mathcal{X} + \{(0,0) \}$. So we are done.

But if $S_1$ is open, then $\bar{S_1} = S_1 \cup \{(0, t) : t \geq 0 \}$. So, from (Equation 2) we again conclude that:

$\bar{\mathcal{K}}_\mathcal{X} = S_1 \cup \{(0, t) : t \geq 0 \} \cup \{(0, t) : t \geq 0 \} = S_1 \cup \{(0, t) : t \geq 0 \}$

$ \bar{\mathcal{K}}_\mathcal{X} = \{(x,t) : t \geq \frac{||x||}{M}, \frac{x}{t} \in \mathcal{X}' \} \cup \{(0,t) : t \geq 0 \} = \mathcal{K}_\mathcal{X} + \{(0,0) \}$

Answer 1

First note that if $x \in X$ and $t>0$ then ${1 \over n} (x,t) \in X$, so it is clear that $(0,0)$ is in $\overline{K_X}$. And so $K_X \subset C=K_X \cup \{(0,0)\} \subset \overline{K_X}$.

We can write $K_X = \{(x,t) | t >0, x \in t X \}$ and $C=\{(x,t) | t \ge0, x \in t X \}$, so if we can show that $C$ is closed then it must be the closure.

Now suppose $(x_n,t_n) \in C$ and $(x_n,t_n) \to (x,t)$. If $t>0$, then ${x_n \over t_n} \to {x \over t} \in X$ (since $X$ is closed) and hence $(x,t) \in C$. If $t=0$, then since $X$ is bounded, we must have $x = 0$ and so $(x,t) \in C$.

Hence $C$ is closed and so $C=\overline{K_X}$.