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I very well know that every open ball is an open set. and that every open set need not be an open ball. But illustrate me some counter example.

Vijay
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    A counter example on what? An open set that is not a ball? – Marra Sep 26 '15 at 12:17
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    If I get your question, then just look what is the union of 2 open balls with no intersection. And in general open balls are some special cases of open sets. – Mesmerized student Sep 26 '15 at 12:19
  • Take, for example, the subset ${ (x,y)\in\mathbb{R}^2 : |x|<1,|y|<1}$. It's the interior of the square of side size 2, centered at the origin of $\mathbb{R}^2$. It's possible to prove that this set is open and it is obviously not a ball. – Marra Sep 26 '15 at 12:19

2 Answers2

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Here is a counterexample.

Consider the set $\Bbb R$ of real numbers. Let $B(0, 2)$ represent the ball of radius $2$ around $0$, i.e, the interval $(0-2, 0+2) = (-2,2)$.

Now let $B(5,2)$ represent the ball of radius $2$ around $5$, i.e., the interval $(5-2, 5+2) = (3,7)$.

The union of these two balls, $B(0,2) \cup B(5,2)$, is open, but it is not itself a ball because every open ball in $\Bbb R$ can be expressed as an open interval, but how do you express that union as a single open interval? If it helps, feel free to draw a picture of the real line $\Bbb R$ and the intervals I mentioned.

layman
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The most simple examples of open sets, which are not balls, in every metric space are $\emptyset$ and the space itself, which are open.

Even if you consider those sets to be balls with radius $0$ or $\infty$, respectively, you can take the union of two or more open balls, which is not necessarily an open ball anymore (see user46944's answer for the one-dimensional case).

What you can actually show instead is that every open set in a metric space can be written as a union of open balls (see this quesion). In $\mathbb{R}^n$ you can furthermore show that a set is open if and only if it is a countable union of open balls (see this question).

mkausp
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