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This is a follow up question to the one I posted here. So my new question involves accumulation points and infinity. Can infinity be considered an accumulation point? For example consider the following set $$S = \{ 2^n + \frac{1}{k} \space : \space n, k \in N \}$$ Now if we fix $n$ the set tends to $2^n$ but if we fix $k$ the set tends to $\infty$. Does this mean that accumulation points exist at $2^n$ and $\infty$? Any help would be appreciated.

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  • $\begingroup$ If you think of $\infty$ as defining the one-point compactification of the set, then sure. $\endgroup$ – Euler....IS_ALIVE Feb 6 '13 at 3:32
  • $\begingroup$ Yes. If you know the definition of accumulation point, this should be straightforward to show. Furthermore, these are the only accumulation points. $\endgroup$ – Brett Frankel Feb 6 '13 at 3:32
  • $\begingroup$ There's a confusing typo in the sentence that starts with "Now if". $\endgroup$ – Michael Hardy Feb 6 '13 at 3:39
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    $\begingroup$ It depends on the ambient set and topology. If it is $\mathbb{R}$ with the usual topology, then $\infty$ is not a point, so cannot be an accumulation point. So the accumulation points are $\{2^n\}_{n \in \mathbb{N}}$ in this case. $\endgroup$ – copper.hat Feb 6 '13 at 3:41
  • $\begingroup$ I've applied what I think is the appropriate correction to the typo @MichaelHardy mentions, based on context; please let me know if this is wrong. $\endgroup$ – Steven Stadnicki Feb 6 '13 at 3:54
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When thinking about limits and accumulation points one needs to really think topologically and keep in mind the the names of the points in your space are just names, nothing more. If your space does not contain a point named $\infty$ then infinity can't be an accumulation point of anything. So that would be the first thing to notice. Now, there are various different and useful ways of adjoining infinit(y)(ies) into familiar spaces. Above you are interested in a subset of $\mathbb R$ (that certainly has all $2^n$ as accumulation points as you mention) so you need to consider ways of adjoining something you'd like to call 'infinity' into the space $\mathbb R$. One way of doing that is to add two infinities, namely $\infty $ and $-\infty $. Then you need to extend the topology to include these.

There are various natural ways of doing that, and they will lead you to construct a topology on $\mathbb R \cup \{\pm \infty \}$ that is homeomorphic to $[-1,1]$ with the usual topology. With that topology $\infty $ will be an accumulation point of your set. Equivalently, applying the homeomorphism to the set will give you a subset of $[-1,1]$ that has $1$ as an accumulation point. This illustrates well that the names of the points are just names, nothing more, and now $\infty $ has been completely demystified.

Another possibility of adding infinity to $\mathbb R$ is to merge $\infty $ and $-\infty $ into a single infinity. The result space is going to be homeomorphic to a circle, say $\{z\in \mathbb C \mid |z|=1\}$. This can be done in such a way that the homeomorphism maps the new infinity to any point on the circle you like, showing again the insignificance of the names of the points in the space.

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The answer to the question is yes.

If you're talking about $+\infty$, as opposed to $-\infty$ and also as opposed to a single $\infty$ at both ends of the line, then a basic open neighborhood of $+\infty$ is a set of the form $$ (a,+\infty] = \left\{ x\in\mathbb R\cup\{+\infty\} : x>a \right\} = \left\{ x\in\mathbb R\cup\{+\infty\} : (x\in\mathbb R\ \&\ x>a)\text{ or }x=+\infty \right\}. $$ And an open neighborhood of $+\infty$ is any open set that has as a subset some basic open neighborhood of $+\infty$.

The point $+\infty$ is an accumulation point of a set $A$ precisely if every open neighborhood of $+\infty$ intersects $A$ at at least one point other than $+\infty$.

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