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Let us take the connected metric space $(\mathbb{R},d)$ where $d(x,y)$ may be defined as $d(x,y)=|x-y|$. Now let us define the subset $\mathbb{N}$ of $\mathbb{R}$. No matter how we choose $r$, every neighborhood $N_r(x)$ where $x$ is a natural number, would contain all real numbers in $(x-r,x+r)$, and hence would not be entirely contained in $\mathbb{N}$. Hence no point in the subset $\mathbb{N}$ of $\mathbb{R}$ can be an interior point. By this logic it seems to me that if we define a subset of countable points chosen from a connected metric space, it would not have any interior points. Is that correct?

Canine360
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    What is a "continuous" metric space? What is a subset "defined over" a metric space? –  Jul 30 '16 at 01:43
  • Yeah you're right, I have used imprecise terminology. "Defined over" is not the right phrase, it would be "subset of". And "continuous metric space" can be replaced with a metric space (X,d) where X has uncountably infinite number of elements (and the subset would consist of a countable set of points chosen from this set). Just like N (subset) and R (metric space), as I've mentioned. – Canine360 Jul 30 '16 at 01:46
  • For example, $[0,1] \cup {2}$ (with the usual metric of $\mathbb R$) is an uncountably infinite metric space, and the discrete subset ${2}$ has an interior point, namely $2$. – Robert Israel Jul 30 '16 at 01:53
  • Have edited the question to (hopefully) make the terminology more precise, as suggested in comments. – Canine360 Jul 30 '16 at 01:55
  • OK. Maybe I should say uncountable metric space with no discontinuity (I'm not sure if discontinuity is the right word, as mathguy pointed out), because ${2}$ is again a discrete point which has been added to a continuous interval $[0,1]$ to make this metric space. Hence in this metric space the point 2 doesn't behave any differently (in terms of neighborhoods etc.) from the way it would if my metric space consisted of natural numbers only. You get the idea of what kind of "continuous" metric space I'm talking about. Can anyone help me with more precise terminology maybe? Thank you. – Canine360 Jul 30 '16 at 02:06
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    @Canine360 You might be looking for the notion of a connected metric space (i.e. one that cannot be partitioned into two non-empty open sets). It is true in this case. Or you might ask that it have no isolated points (i.e. there is no point $x$ in the metric space such that there exists a $\varepsilon>0$ such that $d(x,y)>\varepsilon$ for all $y\neq x$). The answer is "yes" in the first case and "no" in the second (for instance, $[0,1]\cup \mathbb Q$ is a counterexample then) – Milo Brandt Jul 30 '16 at 02:15
  • @Milo Brandt: Thanks so much. Connected metric spaces is exactly what I was looking for. I wasn't aware of the concept of connected metric spaces, hence was not able to put my finger on it. (It has not been defined in my textbook, Rudin, till whatever part I've read.) Looked it up after you mentioned it. Thanks again. – Canine360 Jul 30 '16 at 02:21
  • Have edited the question again to include connectedness, as suggested. Thanks again everyone. – Canine360 Jul 30 '16 at 05:54

2 Answers2

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Suppose $A$ is a countable subset of a connected metric space $M$, and $x\in M$; we'll show $x$ is not an interior point of $A$.

Towards contradiction, suppose $x\in int(A)$. Then there is some positive $\epsilon$ such that $B_\epsilon(x)\subseteq A$. But since $A$ is countable, and the interval $(0, \epsilon)$ is uncountable, there must be some $\delta<\epsilon$ such that $d(x, a)\not=\delta$ for any $a\in A$.

Since $B_\epsilon(x)\subseteq A$ and $\delta<\epsilon$, this means in fact that $d(x, y)\not=\delta$ for any $y\in M$.

But now we can write $M=I\sqcup O$, where

  • $I=\{y\in M: d(x, y)<\delta\}$, and

  • $O=\{y\in M: d(x, y)>\delta\}$.

Each of $I$ and $O$ is open, so this contradicts the connectedness of $M$.

Noah Schweber
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As posted, the answer is "NO." Here is an example. Take
$ X = \mathbb{R}\times \{0\} \cup \mathbb{N}\times \{1\} $
i.e.: $X$ is the x-axis in $\mathbb{R}^2$, union $\mathbb{N}$ at height 1. Clearly $X$ is uncountable, and with the usual distance in $\mathbb{R}^2$, all points on the countable subset $\mathbb{N}\times \{1\} $ are interir points, since for any $ n \in \mathbb{N} $ , the ball centered at $p=(n,1)$ of radious 0.5, which reduces to the set $\{p\}$, is contained in $\mathbb{N}\times \{1\} .$

VictorZurkowski
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  • I think the answer proves the opposite, i.e. it is true (for connected sets, which is clarified in question comments). – Canine360 Jul 30 '16 at 05:51