# Can an open ball have just one point. As per my understanding it cannot. Please clarify.

I am new to functional analysis and am just learning. To my understanding an open ball must have at least 2 points else its definition will not be satisfied.

Now if I have just an empty set and this open ball, why cannot it constitute a Topology. I see it satisfying intersection and union conditions.

I agree that the question is too basic!

Definition of open ball : $$B_r(x)= \{y \in E | d(x,y)<r \}$$ $(E,d)$ is the metric space.

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You say "else its definition will not be satisfied". Can you tell us what definition you are thinking of? I usually think of an open ball as something that would be part of a metric space---there needs to be a "radius" for the "ball". In that sense, this would be about something more specific than an abstract topology. –  alex.jordan Feb 12 '14 at 0:20

In metric spaces, open balls may have just one point: Let $X$ be any (non-empty) set and consider the following metric in $X$: $d(x,y)=\begin{cases}0&,if\ x=y\\1&,if\ x\neq y\end{cases}$. Then the open ball of radius $1/2$ centered at any $x\in X$ is simply $\left\{x\right\}$. For a less artificial example, take $Y=[0,1]\cup\left\{17\right\}$ with the metric induced from $\mathbb{R}$. Then the open ball (in $Y$!) centered at $17$ with radius $4$ is just $\left\{17\right\}$.

On the other hand, if you have a non-trivial normed space $X$ (over $\mathbb{R}$, say), then any open ball in $X$ will have infinite points.

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But as per the definition of distance which I just learnt, if x=y, then $d(x,y)=0$ for this to be a valid distance metric. Am I wrong? –  kosmos Feb 12 '14 at 0:29
Oops. Sorry, I wrote it wrong. Thank you. –  Luiz Cordeiro Feb 12 '14 at 0:31
@kosmos That's rught. But if you have the metric space as in Luiz's example (called the discrete space), then for any $x\in X$, the only point $y\in Y$ satisfying $d(x,y)<1/2$ is $x$ itself! More generally, the open ball $d(x,y)<r$ consists only of $y=x$ for $r\leq 1$, or the whole space for $r>1$. –  triple_sec Feb 12 '14 at 0:32
@kosmos The ball $B_{1/2}(x)$ is defined to be the set $\{y \in X \mid d(x,y) < 1/2\}$. Since $d(x,x) = 0 < 1/2$, we know $x \in B_{1/2}(x)$. –  Austin Mohr Feb 12 '14 at 0:39
Thanks @AustinMohr –  kosmos Feb 12 '14 at 0:49

You say that you are studying functional analysis, so perhaps you are mainly interested in Banach spaces. But even in that context, you are only almost correct - any open ball of a non-trivial Banach space is infinite. The trivial Banach space $V=\{0\}$ consists of just its zero vector $0$, and thus for any $r>0$, $$B_r(0)=\{v\in V:|v-0|<r\}=\{v\in \{0\}:|v|<r\}=\begin{cases} \{0\}&\text{if }|0|<r,\\ \varnothing&\text{if }|0|\geq r \end{cases}=\{0\}=V,$$ which is therefore both an open ball and a singleton set.

Of course, most metric spaces are not Banach spaces, and there are many other counterexamples. Basically, for any metric space $X$, if $x\in X$ is an isolated point, then $\{x\}$ will be an open ball of $X$.

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Have not reached Banach spaces yet. :( Will read this again when I do. –  kosmos Feb 12 '14 at 0:30

It depends on the space. In $\mathbb{N}$ with the usual topology inherited from the real line, the open 1-ball about any point is just that point. Such points are said to be isolated.

Your example is correct. It is a trivial space consisting of a single point, and you have observed the only topology that exists for that space.

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