I would like some help clarifying the definition of open balls in the discrete metric space.

The definition I am provided is:

Open balls in the discrete metric space $M = (X,d_0) $ are given by $B_\epsilon(x) = \left\{\begin{matrix} \{x\} & \epsilon \leq 1\\ X & \epsilon > 1 \end{matrix}\right. $

My question:

Why is it not this way: $B_\epsilon(x) = \left\{\begin{matrix} \{x\} & \epsilon < 1\\ X & \epsilon \geq 1 \end{matrix}\right. $

My reasoning is because the points lying on the boundary do not lie in $B_\epsilon(x)$, and should be considered NOT part of it. Or am i missing something?

Many thanks.

  • $\begingroup$ If $y\neq x$ then $d(x,y)=1$ hence $y\notin B_1(x)$ ($d(x,y)=1$ is not $< 1$) this is why the first works. For the boundary, for the discret topology, the boundary of any set is itself (every set is clopen). In particular every open ball being also closed is its own boundary. $\endgroup$ – Clément Guérin Apr 23 '15 at 11:58

Your reasoning is correct. Think about $\mathbb R^2$ with the usual euclidean metric. An open ball does not contain its boundary.

In general, the open ball $B_r(x_0)$ in a metric space $(X,d)$ is defined to be $$B_r(x_0) := \{x \in X : d(x_0,x)<r\}$$

  • $\begingroup$ Thank you, exactly what you said yes - in line with the definition. I guess there is a typo in the book then ... $\endgroup$ – beginner Apr 23 '15 at 11:56
  • $\begingroup$ @SanjouOdawali No, the book is correct. If you put $r=1$ then $B_1(x) = \{x\}$. $\endgroup$ – GenericNickname Apr 23 '15 at 12:03
  • $\begingroup$ Thank you Sir, I get it now. $\endgroup$ – beginner Apr 23 '15 at 12:26
  • $\begingroup$ Hi, I am so sorry! apparently I am still a little lost. Suppose we use B1(1) on the real line (i.e. open ball centered at 1 with radius 1) Using the definition (< 1), then yes agreed that the open ball is simply {1}. But if it is <= 1 as per the above definition in the question, wouldn't the open ball it be {0,1,2} ? $\endgroup$ – beginner Apr 23 '15 at 12:58
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    $\begingroup$ What you have posted above is no definition. Its the result of applying the definition of an open ball in the special case of a discrete metric. The "$\leq$" in your question does NOT mean that you take all elements around the center which have distance of $\leq \varepsilon$! It states what the balls look like for different radii. And if you take radius $=1$, you still consider only elements with distance strictly lesser than $1$! $\endgroup$ – GenericNickname Apr 23 '15 at 13:13

You are right. By the very definition of $\def\eps{\varepsilon}B_\eps(x)$, we have $$ B_\eps(x) = \{y \in X : d(x,y) < \eps\} $$ hence $$ B_1(x) = \{y \in X : d(x,y) < 1 \} $$ Now the points with distance $1$ to $x$ (that is all points but $x$), do not belong to $B_1(x)$.

  • $\begingroup$ Thank you for clarifying martini, the definition is indeed the way to go $\endgroup$ – beginner Apr 23 '15 at 11:57

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