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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

3 Answers 3

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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 $(0,1]$.

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    Okay, your last example turned on some lights. So, even though I could choose some $x \in (0, 1]$ and $r > 0$ for which $1 \in B_r(x)$, the fact that $x$ isn't 1, and that the statement doesn't hold for $x = 1$, shows that the set is not open. – 2017-02-01
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    Right. The set $(0,1]$ is only open if for every $x$, there is an $r > 0$ so that $B_r(x) \subset (0,1]$. For $x \in (0,1)$, there is such an $r$. But for $x =1$, there is no $r > 0$ so that $B_r(1) \subset (0,1]$ – 2017-02-01
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    So, how does that work for a closed ball, then? To show that [0, 1] is closed, what radius would our ball need to be at $B_r(1)$ to still be in the interval? We can't extend our ball outside 1 - I imagine this is where the $||x - x_0|| \leq r$ comes in handy? – 2017-02-01
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    What is your definition of closed? The usual definition is that a set is closed if its complement is open. So to show that $[0,1]$ is closed, you would need to show that $(-\infty,0) \cup (1,\infty)$ is open. – 2017-02-01
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Here's an equivalent, but more interpretable definition of an open set in $\mathbb R^n$.

Say that a set $U \subset \mathbb R^n$ is open if for every $x \in U$, there is some $r > 0$ such that $B_r (x) \subset U$.

Hence, a set is open if, for each point in the given set, you can find an open ball around that point that lies completely in the set.

Can you see why the above definition is equivalent to yours?

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'it seems obvious that I could just choose a ball with an $r$ small enough to contain my point, and be inside $U$ still'

You're right. The fact that you can do this for any point in the set is what makes it open.

Compare to a non-open set, such as a closed ball or a single point. Can you find any points in these examples such that you can't put an open ball around them that remains within the set? (For a set consisting of a single point, there's only one option.)