Not too sure where to start for this question, any help would be appreciated!
The definition of connected that I am using is:
A connected space cannot be written as the disjoint union of two nonempty open sets.
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1Since connected space is a topological property, is there any topology $\mathcal{T}$ on $\mathbb{N}$ which makes $(\mathbb{N}, \mathcal{T})$ a connected space ? – Matthieu Jentile Oct 29 '20 at 22:11
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@MatthieuJentile I don't know what a topology is yet, unfortunately. – spaghettihoops Oct 29 '20 at 22:18
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The usual definition of "connected" requires knowing some of the language of topology. But since, as you say, you don't know topology yet, you should probably edit your question to tell us what definition of connected you are using. – Lee Mosher Oct 29 '20 at 22:29
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@LeeMosher Thanks, updated! – spaghettihoops Oct 29 '20 at 22:32
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Good. So for purposes of reading and understanding the various answers to your question, you should realize that the "topology" of a metric space is nothing more than the collection of its open subsets, and "topological properties" are nothing more than the properties that can be derived by study of the open subsets. And that's what's in those answers, namely proofs about the open subsets. – Lee Mosher Oct 29 '20 at 22:53
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Does this answer your question? Does a connected countable metric space exist? – Oct 30 '20 at 07:22
2 Answers
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Let $d$ be the supposed metric such that $\mathbb{N}$ is connnected. Then $d:\mathbb{N}^2\to\mathbb{R}^+$ is a continuous function. As such, it sends connected sets to connected sets, and so $d(\mathbb{N}^2)$ is connected. However, given that it is at most numerable, and that connected sets in $\mathbb{R}^+$ are intervals, we must have $\#d(\mathbb{N}^2)=1$. Since $d(x,x)=0$, $d(\mathbb{N}^2)=\{0\}$. This is a clear contradiction, and it generalizes to non trivial metric spaces with cardinality less than $\mathfrak{c}$
Note that there's a topology such that $\mathbb{N}$ is connected: the trivial topology, i.e. $\tau_\mathbb{N}=\{\emptyset, \mathbb{N}\}$
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Is there a way to answer the question without using topology? I haven't got that far yet (just learning about metric spaces) – spaghettihoops Oct 29 '20 at 22:27
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@spaghettihoops the answer does not use topology. I just added the final sentence as a marginal note, but it is not relevant to the answer in itself. If you have any particular doubts, feel free to ask – Oct 29 '20 at 22:28
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There are countably infinite connected topological spaces, one of which I described in this answer, but there is no countably infinite connected metric space. To see this, suppose that $d$ is a metric on $\Bbb N$. Let $D=\{d(0,n):n\in\Bbb Z^+\}$; $D$ is a countable set, so there is a positive $r\in\Bbb R\setminus D$. Let $U=B(0,r)$, the open ball of radius $r$ centred at $0$; I claim that $\Bbb N\setminus U$ is also open and hence that $\langle\Bbb N,\rangle$ is not connected.
Let $n\in\Bbb N\setminus U$; then $d(0,n)>r$. Let $s=d(0,n)-r>0$; it’s easy to verify that $B(n,s)\cap U=\varnothing$, so $B(n,s)$ is an open nbhd of $n$ contained in $\Bbb N\setminus U$. And $n$ was an arbitrary element of $\Bbb N\setminus U$, so $\Bbb N\setminus U$ is open.
Brian M. Scott
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Is there a way to answer the question without using topology? I haven't got that far yet (just learning about metric spaces) – spaghettihoops Oct 29 '20 at 22:27
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@spaghettihoops: Connectedness is a topological concept, so there’s no way to avoid topology completely. I gave an argument that stays entirely within the domain of metric spaces. What parts seem completely unfamiliar? – Brian M. Scott Oct 29 '20 at 22:32
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We haven't defined a topological space yet, only metric spaces. – spaghettihoops Oct 29 '20 at 22:37
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@spaghettihoops: That’s okay: you can just ignore the first part of the first sentence. I included it partly for completeness but mostly because of Matthieu Jentile’s comment. Everything else in the answer is metric space concepts. – Brian M. Scott Oct 29 '20 at 22:42