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 April 2nd, 2014, 03:26 PM #1 Member   Joined: Feb 2013 Posts: 80 Thanks: 8 Questions about balls... I am not sure if I made this thread more for typing the thread title or getting answers. Anyways: The metric space, $\displaystyle (\mathbb{N},d)$ for some arbitrary distance function, d, is a discrete metric space--yes? (1) If so, then we continue to define a subset $\displaystyle N$ of any metric space, as a neighbourhood if $\displaystyle \exists \delta > 0$ such that $\displaystyle B(a;\delta) \subset N$. (2) And further then $\displaystyle \forall b \in B(a;\delta)$, $\displaystyle B$ is a neighbourhood of $\displaystyle b$ as well. (3) So back to the previously defined [discrete] metric space, I ask if,"All open balls in $\displaystyle (\mathbb{N},d)$ are also closed."is true. (4) And if it is, then by the idea from (2), all points $\displaystyle b$ in the ball $\displaystyle B(a;\delta)$ have the quality such that $\displaystyle B(a;\delta)$ is a neighbourhood of b. However this brings me to my final question. Does (4) then imply, because of the definition of a neighbourhood in (1), that a ball in this discrete metric space is able to only contain one point at the [closed] edge of the ball, that one point being the point the ball is centred on? Thanks, and I am sorry for my (probably really) confusing line of questioning. It's just with a non-discrete metric space it is easy to understand (2). However, under the conditions above it leads me to the assumption that a ball is able to contain just one point, as on the edge of a closed ball, any $\displaystyle \delta$ that encloses more than just the point on the edge would also enclose a point outside of the original ball, thus not fulfilling (1). Thanks again! April 2nd, 2014, 03:51 PM   #2
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 Originally Posted by anakhronizein Does (4) then imply, because of the definition of a neighbourhood in (1), that a ball in this discrete metric space is able to only contain one point at the [closed] edge of the ball, that one point being the point the ball is centred on?
Yes. For any $\displaystyle n\in\mathbb N$ and $\displaystyle 0<\delta<1$, $\displaystyle \mathrm B(n,\delta)=\{n\}$. April 2nd, 2014, 04:13 PM   #3
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 Originally Posted by Olinguito Yes. For any $\displaystyle n\in\mathbb N$ and $\displaystyle 0<\delta<1$, $\displaystyle \mathrm B(n,\delta)=\{n\}$.
So even though it doesn't contain any other points that enclose $\displaystyle n$, that $\displaystyle \delta$ still is sufficient to say there is a ball? April 2nd, 2014, 04:24 PM #4 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra That’s right. In $\displaystyle \mathbb N$ the distance between any two distinct points is at least 1 so a ball with radius less than 1 would contain just one element, the element it is centred on. Thanks from MarkFL and anakhronizein April 7th, 2014, 08:15 AM #5 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 A few observations about this space: Of course, for balls of "sufficiently large radius" we have other points in a neighborhood of $n$ besides $n$, but the point in topology is studying "local conditions" (the neighborhoods we are interesting in are "small"). Every singleton is indeed "clopen" (open and closed)...this expresses the idea that a discrete space is totally disconnected. In other words, the only topology a metric induces is the full power set of $\Bbb N$, since because singletons are open, every non-empty set is open, and since we have more than one singleton, the empty space is open as well (since we require topologies to be closed under finite intersections). So, in a discrete space like this, we have: $U$ is a neighborhood of $n$ if and only if $n \in U$, for any set $U$. Topologically, this should be viewed as a kind of "default" topology, it doesn't tell us ANYTHING about spatial relationships other than "different points don't touch each other". We can induce such a topology on ANY set, which tells us absolutely nothing specific. It's as if "each point lives in it's own universe". USUALLY, when we study a metric space, we are interesting in "things" that act like "multi-dimensional continua" (such as a plane, or a smooth surface, more generally), that is, the metric function is surjective. With such beasts, even though neighborhoods CAN be "large", a "generic" neighborhood is points CLOSE to a given point. The metric answers the question "what does close mean?" (the answer usually being: "less than epsilon"). Thanks from MarkFL, anakhronizein and Olinguito April 9th, 2014, 04:55 PM   #6
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 Originally Posted by anakhronizein I am not sure if I made this thread more for typing the thread title or getting answers.
I'm sure you'll consider this a laff riot then Jacques Tits - Wikipedia, the free encyclopedia April 9th, 2014, 05:08 PM   #7
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 Originally Posted by Maschke I'm sure you'll consider this a laff riot then Jacques Tits - Wikipedia, the free encyclopedia
Known for the "Tits alternative". Never would have thought that mathematics would fill the need for that! Tags balls, discrete, metric, questions, topology ### questions on felta neighbourhood

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