Showing that the set $\left\{p \in \mathbb{R}_+^L: \sum_{l=1}^{L} p_l =1, p_l >0, \forall l\right\}$ is open Let $\Delta = \{p \in \mathbb{R}_+^L: \sum_{l=1}^{L} p_l =1\}$ where $\mathbb{R}_+^L = \underbrace{\mathbb{R}_+ \times \cdots \times \mathbb{R}_+}_{\text{L times}}$ and $\mathbb{R}_+ = [0, \infty)$, $p_l$ denotes the $l^{\text{th}}$ component of $p$. Let $\Delta^0 = \{p \in \Delta: p_l >0, \forall l\}$. Prove that $\Delta^0$ is an open set in $\Delta$.
Using the definition of an open set, I need to show that for all $p \in \Delta^0$, there exists $\epsilon>0$ such that $\{x \in \Delta: ||x-p|| <\epsilon\} \subseteq \Delta^0$. I am stuck at this step, how can I show the existence of this $\epsilon$?
 A: $\Delta^0$ is not open in $(\mathbb{R}_+)^L$. I'm assuming that $\|\cdot\|$ is the Euclidean norm here.
Let $x_0 = \left(\frac1L, \ldots, \frac1L\right) \in \Delta^0$. Let's show that there does not exist an open ball $B(x_0, r)$ such that $B(x_0, r) \subseteq \Delta_0$.
Indeed, if such $r > 0$ would exist, you would have $\left(\frac1L + \frac{r}2,\frac1L, \ldots \frac1L\right) \in B(x_0, r)$ since:
$$\left\|x_0 - \left(\frac1L + \frac{r}2,\frac1L, \ldots \frac1L\right)\right\| = \left\|\left(\frac{r}2,0, \ldots 0\right)\right\| = \frac{r}2 < r$$
But, $\left(\frac1L + \frac{r}2,\frac1L, \ldots \frac1L\right) \notin \Delta^0$ since
$$\frac1L + \frac{r}2 + \underbrace{\frac1L + \cdots + \frac1L}_{n-1} = 1 + \frac{r}2 > 1$$
This is a contradiction.

To answer the updated question, let's show that $\Delta^0$ is open in $\Delta$. 
Let $x = (x_1, \ldots, x_L) \in \Delta^0$ be arbitrary. Denote $r = \min\{x_1, \ldots, x_L\}$. Now we have $B_\Delta(x,r) \subseteq \Delta_0$.
Indeed, let $y = (y_1, \ldots, y_L) \in B_\Delta(x, r)$. For any $i \in \{1, \ldots, L\}$ we have $$x_i = x_i - y_i + y_i \le |x_i - y_i| + y_i \le \|x - y\| + y_i < r + y_i$$
Hence, $y_i > x_i - r \ge 0$ so $y \in\Delta^0$.
Therefore, $\Delta^0$ is open in $\Delta$.
