I need help with the intuition behind applying open and closed balls on a subset to determine if it is open or closed. By balls, I mean the open ball $B_r(x_0) = \{x \in \mathbb{R}^d : ||x – x_0|| < r\}$ and closed ball $\overline{B_r}(x_0) = \{x \in \mathbb{R}^d : ||x – x_0|| \leq r\}$
Our definition is that if $U \subset \mathbb{R}^d$, we say $U$ is open if $\forall x \in U \space \space \exists B_r(x) \subset U$
I struggle with what this is exactly saying. Let's say $U$ is an open set; our open ball is centered at an $x \in U$, and then our open ball contains the $x \in U$ and also, because we can only have $x \in \mathbb{R}^d : ||x – x_0|| < r$, any $x \in \mathbb{R}^d \cap U$?
I don't know. I'm struggling to put the pieces together in a coherent way – especially because it seems obvious that I could just choose a ball with an $r$ small enough to contain my point, and be inside $U$ still.
Any help on how to think about this would be greatly appreciated. Don't be afraid to point out what I got wrong in my analysis above – I know I'm obviously getting something wrong! haha
Best Answer
You don't have a very clear question, but I think it's best to think about some examples.
First, let $U = B_r(x)$, for $x \in \mathbb{R}^d$ and $r > 0$. We use the word "open" when talking about this ball, but is it really open? Well, yes. Any $y \in B_r(x)$ is contained in a ball contained in $U$. Namely, we can always choose the ball $B_r(x)$. That's why we call them open balls.
In contrast, the closed ball $\overline{B}_r(x)$ is not open. Why? Well, think of a point $y$ on the boundary. There is no $r > 0$ so that $B_r(y)$ is contained in $\overline{B}_r(x)$. Any nonzero radius ball centered at $y$ will intersect the complement of $\overline{B}_r(x)$.
Here's another example. Let $U$ be the upper half plane $\{(x,y) \in \mathbb{R}^2: y > 0\}$. Is $U$ open? Yes. Let $(x,y) \in U$. Then the ball $B_{y/2}(x,y)$ is contained in $U$, since it stays entirely above the $x$-axis. So any $(x,y) \in U$ has an open ball containing it that is itself contained in $U$.
What this definition of openness does is weed out sets that have boundary and isolated points. Take the set $(0,1) \cup \{2\}$. It's not open because $\{2\}$ doesn't have any ball around it contained in our set. Likewise, $(0,1]$ is not open because the point 1 doesn't have any ball around it contained in $(0,1]$.