# Finding the closure of a subset

I have that problem:

We have $(\mathbb{R}^2,\tau)$, where $\tau$ is the standard topology. Find the closure of $$A = \{ (x,y)\in\mathbb{R}^2\ \ |\ \ x^2+y^2<1 \}$$

I know that the boundary of A is $A'= \{ (x,y)\in\mathbb{R}^2\ \ | \ \ x^2+y^2=1 \}$, and I know $\overline{A}=A\cup A'$.

What's the faster and simple way of solve that problem? Is it probing that $A'$ is a set of limit points and probing that there is no more limit points?

• For problems like this, 'guess & verify' is a good approach. Commented Aug 18, 2014 at 17:53

Here's one way of solving it, although you'll have to assess if it qualifies as "faster and simpler".

Note that $B := \{(x,y): x^2+y^2\leq 1\}$ is closed in $\mathbb R^2$ and contains $A$, so $\overline{A} \subset B$.

Conversely, any for any element $(x,y) \in B - A$, the sequence in $A$ defined by $$(\frac{x}{1+\frac{1}{n}}, \frac{y}{1+\frac{1}{n}})$$ converges to $(x,y)$. Closed sets contain all their limit points, so any closed set containing $A$ must contain each element in $B$, and $B \subset \overline{A}$.

Notice this argument does come down to "probing that there are no other limit points", as you suggest: this is exactly what we are doing when we bound $\overline{A}$ with $B$ in the first part.

• I can't really understand this solution. I'm not following you since the sequence until $B\subset\overline{A}$. Can you help me to understand it? Commented Aug 19, 2014 at 6:51
• The basic idea is to prove that $B =\overline{A}$ by showing that $B \subset \overline{A}$ and $\overline{A} \subset B$. To show $\overline{A} \subset B$: recall that $\overline{A}$ is the intersection of all closed sets containing $A$, so if $B$ is closed and contains $A$ then $\overline{A} \subset B$. To show $B \subset \overline{A}$: since $A$ is a subset of its closure, we only need to check that elements of $B-A=\{p \in B: p \notin A\}$ are in $\overline{A}$. Commented Aug 19, 2014 at 13:43
• Note that $B-A=\{(x,y): x^2+y^2=1\}$, which you call $A'$. For any $(x,y) \in B-A$, we have a sequence $\{a_n\}_{n \in \mathbb N}$ defined by $a_n=(\frac{x}{1+\frac{1}{n}},\frac{y}{1+\frac{1}{n}}).$ Notice that (1) $a_n \in A$ for all $n$ and (2) $a_n \to (x,y)$ as $n \to \infty$. If a set is in $\mathbb R^2$ is closed is closed, then it must contain all limits of sequences. We have just shown that each point of $B-A$ is the limit of a sequence in $A$, so any closed set containing $A$ also contains $B-A$. Since $\overline{A}$ is closed containing $A$, this shows $B-A \subset \overline{A}$. Commented Aug 19, 2014 at 13:44
• I hope this helps -- let me know if you'd like further clarification. Commented Aug 19, 2014 at 13:45
• I already understand it and I love your solution. Thanks. Commented Aug 20, 2014 at 3:01

In a normed vector space $E$ if $B(a,r)_o=\{x\in E \, | \, \|x-a\|<r \}$ ($r>0$), then $\overline{B(a,r)_c}=B(a,r)_c=\{x\in X \, | \, \|x-a\|\leq r\}$.
In your example $A=B(0,1)_0$ hence $\overline{A}=\{(x,y) \, | \, x^2+y^2\leq 1\}$.

• What do the subscripts $_c$, $_o$ mean? Commented Aug 18, 2014 at 17:55
• o for open ball, and c for closed ball. Commented Aug 18, 2014 at 17:56
• There is a difference between the closer of the open ball and the closer ball (in metric space), hence the closer ball is not in general $\overline{B(a,r)}$. Commented Aug 18, 2014 at 18:03
• For example in $\Bbb N$ with the standard metric, the open ball $B(0,1)$ is exactly the set $\{0\}$ it's closer is itself. But the closed ball is $\{n\in \Bbb N\ ,\ |n|\leq 1\}=\{0,1\}$. Commented Aug 18, 2014 at 18:08
• I'm not sure if I can use that answer in my Topology subject. Maybe I have to define a metric $d(x,y)=\|x-y\|$ instead of saying "In a normed vector space" Commented Aug 18, 2014 at 18:23