# Showing a point is not a open set in $\mathbb{R}^n$ but a open ball is a open set in $\mathbb{R}^n$

I know this is a common example in most text books and is probably a duplicate question, but I wanted to do it a little differently. The only definition I want use is: A set is S open in if for each x in S there is an open ball around x contained in S. So I don't want to use the triangle inequality of a metric or any real analysis definition of open sets.

1) Let $x \in \mathbb{R}^n$ and we want to show $\{x\} \subset \mathbb{R}^n$ is not open set in $\mathbb{R}^n$. Suppose $\{x\}$ is a open set in $\mathbb{R}^n$. Since the set only has one element, than $B(x,r) =\{y\in \mathbb{R^n}:d(x,y) < r\} \subset \{x\}$. Than $B(x,r) = \{x\}$ but I am having trouble why this is a contradiction since for $r>0$, $d(x,x) < r$.

2) Let $x \in \mathbb{R}^n$, we want to show $B(x,r) =\{y\in \mathbb{R^n}:d(x,y) < r\}$ is a open set in $\mathbb{R}^n$. For any $y \in B(x,r)$, take $r^* = r - d(x,y)$. Now we want to show $B(y,r^*) \subset B(x,r)$, I know we can use the triangle inequality of a metric, but I want to avoid using that and do this some other method, I tried contradiction. Let $z \in B(y,r^*)$, now assume $d(x,z) > r$ If we draw a diagram of this situation, its clear to see that $d(y,z) > r^*$ and we would have a contradiction, but I am having trouble arriving at $d(y,z) > r^*$ from the assumption $d(x,z) > r$.

• Your definition is badly phrased. A set is $S$ open if for each $x$ in $S$ there is an open ball around $x$ contained in $S$. – Pedro Tamaroff Sep 23 '13 at 2:26
• @PeterTamaroff Thanks let me edit. – user77404 Sep 23 '13 at 2:27
• Also, you're working in a metric space, so you have to use the definition of metric space. This includes the triangle inequality. It is pretty strange to want to prove an open ball is open without that. – Pedro Tamaroff Sep 23 '13 at 2:27
• I know, but in my course we haven't introduce the idea of a metric space yet, we are only told this definition to work with. – user77404 Sep 23 '13 at 2:28
• Well, but the triangle inequality still holds in $\Bbb R^n$ with the usual distance. – Pedro Tamaroff Sep 23 '13 at 2:29

The singleton $\{x\}$ is not open because for any $\epsilon >0$, $B(x,\epsilon)$ contains more than one point, so it cannot be contained in $\{x\}$. In particular, if we set $x_k=x+\frac 1 k$ (in the sense we sum $1/k$ to each coordinate) we can take $k$ large enough so that $$|x-x_k|=k^{-1} n^{1/2}<\epsilon$$ For the second one, your approach is fine, just use the triangle inequality which is true in $\Bbb R^n$ regardless of you knowing the definition of what a metric space is.
Let $y\in B(x,r)$. Define $r'=r-d(x,y)$ as you did. Let $B'=B(y,r')$. Pick $x'\in B'$. Then $d(x',y)<r'=r-d(x,y)$. Thus $$d(x,x')\leqslant d(x,y)+d(y,x')<d(x,y)+r-d(x,y)=r$$ so that $x'\in B(x,r)$ and $B(y,r')\subseteq B(x,r)$, so $B(x,r)$ is open.
• I understand intuitively we can find another point in $B(x, r)$, but to be a little more formal can we explicitly find this point in $\mathbb{R^n}$, so we could make a argument like for r > 0, $y = ...$ will be in $B(x, r)$? – user77404 Sep 23 '13 at 2:36
• The solution to second question is clear, thanks. But what I meant to say in my previous comment is for the first question how can we explicitly find $y \in B(x, \epsilon)$ such that y is not x. Intuitively I know its clear that there will be another point in $B(x, \epsilon)$. – user77404 Sep 23 '13 at 2:45