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.

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    $\begingroup$ This set is neither open nor closed. $\endgroup$ – DHMO Apr 14 '17 at 16:13
  • $\begingroup$ How did you figure that out in seconds?! :) $\endgroup$ – TimelordViktorious Apr 14 '17 at 16:13
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    $\begingroup$ 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. $\endgroup$ – DHMO Apr 14 '17 at 16:14
  • $\begingroup$ 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. $\endgroup$ – TimelordViktorious Apr 14 '17 at 16:15
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    $\begingroup$ 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 $\{(x,y) \in \mathbb{R^2}: (x,y) = (0,0)\}$. $\endgroup$ – 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.

As a general rule, when you're showing something does not have a property, its good to be concrete when constructing your counterexample, rather than abstract.

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.

Alternatively, since you know the closure is not equal to the set, you could probably just cite that (if this is for a course). Similarly for the interior.


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