# Is this set not open and not closed?

Let $X = \{(x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1, (x,y) \neq (0,0)\}$

Is this set open, closed, or neither, and what is the boundary, interior and closure?

I want to get a better understanding if I am approaching these problems correctly.

Is this set open? If I take a point in the set X, so that $p = (p_x, p_y)$, then $(p_x, p_y) \neq (0,0)$, and I define an open ball $B_r(p)$, and I take a point $q \in B_r(p)$. Is it possible for $q \in X$? Well no, since if $p$ satisfies $x^2 + y^2 = 1$ then it is possible for q to be outside of the set or that q satisfies $x^2 + y^2 > 1$.

So it is not open.

Is this set closed? $X$ is closed if $X^c$ is open. But $X^c = \{(x,y) \in \mathbb{R^2}: x^2 + y^2 > 1, (x,y) = (0,0)\}$ is not open, since (x,y) = (0,0) defines a boundary point (and set is only open if for every $x \in X^c$, $x$ is an interior point).

Therefore it is neither closed nor open.

Boundary: $\{(x,y) \in \mathbb{R^2}: x^2 + y^2 = 1\}\cup\{(x,y) \in \mathbb{R^2}: (x,y) = (0,0)\}$

Interior: $\{(x,y) \in \mathbb{R^2}: x^2 + y^2 < 1, (x,y) \neq (0,0)\}$

Closure: $\{(x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1\}$

I am not sure if I am thinking about these problems correctly. I am having a lot of trouble, so I could use some help if there is something about my work that is not correct or clear.

• This set is neither open nor closed. – DHMO Apr 14 '17 at 16:13
• How did you figure that out in seconds?! :) – TimelordViktorious Apr 14 '17 at 16:13
• It is not closed because it does not contain all of its limit points. It is not open because it is not its interior. I figured that out in seconds by having a picture in my head. Intuition is sometimes useful for dealing with well-behaved sets like these. – DHMO Apr 14 '17 at 16:14
• Actually that's the line of reasoning I had in my head, but I wanted to iron it out further. Good to know I'm on the right track then. Thank you. – TimelordViktorious Apr 14 '17 at 16:15
• All your answers are correct (apart from the typo that I fixed for you). So you seem to have a pretty good grasp of the concepts. But I suggest simply writing $\{(0,0)\}$ instead of $\{(x,y) \in \mathbb{R^2}: (x,y) = (0,0)\}$. – TonyK Apr 14 '17 at 16:24

Well, when you argue it is open, you say it isn't possible for $q$ to be in the set. This isn't really true. Consider $p=(0.5,0.5)$, which is in the set, and $B_{0.1}((0.5,0.5))$: the $x^2 + y^2$ will certainly be less than $0.6^2 + 0.6^2 = 0.72\le 1$, so every point in this ball is in $X$.
The problem comes from talking about $p$ and $q$ and then choosing them to have certain properties, that they don't necessarily have in general.
I would suggest, when saying something is not open, to consider a particular point: say $(0,1)$, which is certainly in $X$. Any open ball, of radius $r$, around this point will contain the point $(0,1+\frac{r}{2})$, and $0^2 + (1+\frac{r}{2})^2 \ge 1$ so this point is not in $X$. So $(0,1)\in X$ is not an interior point, and $X$ is not open.
Your argument for not-closedness is correct. If you have more machinery about closed sets, you could argue more directly: a closed set should contain all of its limit points, i.e. if $\lim x_n = x$ and all the $x_n$ are in $X$, $x\in X$. Here, we can have a sequence like $(0,\frac{1}{n})$ that goes to $(0,0)$, but $(0,0)$ is not in $X$. I usually find it easier to think about closed sets as closed under limits.