Give an example of a metric space $(X, d)$ and a subset $E \subseteq X$ where $E$ contains an interior point that is not a limit point.

Hint: You won't be able to find an example of $E$ when $X = \mathbb{R}^2$ and $d$ the usual metric. Consider the discrete metric on $\mathbb{R}^2$.

From what I understand, the discrete metric $d(p, q)$ is equal to $1$ when $p \ne q$, and $0$ when $p = q$. In other words, the only "distance" you can have with the discrete metric is either $1$ or $0$, right?

I'm pretty stuck on this problem. Any point $p$ in some subset $E \subseteq \mathbb{R}^2$ can only have one neighborhood, $N_1 (p)$, since $N_0 (p)$ is a singleton, and singletons cannot be neighborhoods. However, it seems for $p$ to have a neighborhood contained in $E$ (i.e. to be an interior point), it needs to be $\ge 1$ "distance" from the boundary of $E$, but then this would also make it a limit point, since all neighborhoods of $p$ will have some point $q \in E$. How should I tackle this problem? Any tips/hints would be great. Thank you.

  • $\begingroup$ There's a trick. "Normally" if $z$ is an interior point of E so that $N(z)\subset X$ then there are $y \in N(z)\subset E$ and if $y \ne z$ then that mean $z$ must also be a limit point of $E$. The only way that can't happen is if there AREN'T any $y \in N(z)$ so that $y \ne Z$. SO that means $N(z) = \{z\}$ (!!!!!). But "normally" that is impossible because $\{n\}$ is normally not an open neighborhood. So ... that's the hint. Find metric space so that $N(z) = \{x \in X| d(x,z) < \epsilon\} = \{z\}$. The discreet space will do that, if $\epsilon < 1$. $\endgroup$ – fleablood Sep 14 '17 at 0:15
  • $\begingroup$ "and singletons cannot be neighborhoods." BINGO!!!!! That is the trick! For this to be true a singleton MUST be a neighborhood. In order for singletons to be neighborhoods you must use a different metric. In the discreet metric, singletons must certainly ARE neighborhoods. EVERY set is a neighboorhood. Hint: Let $N_{\frac 12}(0) = \{x\in X|d(x,z)< \frac 12\} = ?????$. That is a neighborhood by definition. $\endgroup$ – fleablood Sep 14 '17 at 0:20

You write "Any point $p$ in some subset $E\subseteq\mathbb{R}^2$ can only have one neighborhood, $N_1(p)$, since $N_0(p)$ is a singleton, and singletons cannot be neighborhoods." Actually, every point in $\mathbb{R}^2$ has many neighborhoods. In the topology given by the discrete metric, every subset of $\mathbb{R}^2$ is open. Therefore any subset of $\mathbb{R}^2$ will be a neighborhood of each of its points. In particular, the singleton $\{x\}$ is a neighborhood of $x$ because $\{x\}=B(x,1)$, where I am using $B(c,r):=\{y\in\mathbb{R}^2 \mid d(c,y) < r\}$ to denote the open ball centered at $c$ or radius $r>0$.

Let $x\in\mathbb{R}^2$ and set $A:=\{x\}$. Then $x$ is an interior point to $A$ because the open ball $B$ centered at $x$ of radius $1/2$ is equal to $\{x\}$, and hence $x\in B =\{x\} \subseteq A$. However $x$ is not a limit point of $A$ because $B$ is a neighborhood of $x$ which contains no points of $A$ other than $x$; i.e., in symbols, $B\cap (A\setminus\{x\})=\emptyset$.

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  • $\begingroup$ I thought the discrete metric meant that distances can only be 0 and 1. So, how can we have a radius of 1/2? Doesn't that violate the definition of a discrete metric? Clearly there's something I'm not getting, and an explanation would be great. Thanks. $\endgroup$ – Max Sep 13 '17 at 23:49
  • $\begingroup$ @Max define ball in a metric space $\endgroup$ – user223391 Sep 13 '17 at 23:52
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    $\begingroup$ @Max Just because there is no point $y\in\mathbb{R}^2$ distinct from $x$ such that $d(x,y)<1$ doesn't mean we can't consider the open ball $B(x,1/2)$ (or $B(x,1/3)$ or $B(x,1)$ or $B(x,r)$ for $r>0$). It just means that $B(x,1/2)$ happens to be equal to $\{x\}$. In a metric space, we consider a set $A$ to be open if, for every point $a\in A$ there is $r>0$ such that $B(a,r)\subseteq A$. By taking any $0<r\le 1$, we get $B(a,r)=\{a\}$ (which is clearly a subset of $A$ if $a\in A$), and hence we have shown that every subset is open. $\endgroup$ – John Griffin Sep 13 '17 at 23:53
  • $\begingroup$ "I thought the discrete metric meant that distances can only be 0 and 1" Exactly! So if $N(z) = \{d(x,z) < 1/2\}$ then $N(z) = \{d(x,z) = 0\}=\{z\}$ because $0$ is the only acceptable distance less than $1/2$. A radius does not have to be a legitimate distance. A neighborhood with radius 1/2 is all points that have a distance of LESS than 1/2. And $0 \le \frac 12$, isn't it? That does NOT mean that there are any actual points with a distance of 1/2. $d(x,y) = \frac 12$ is impossible. But $d(x,y) =0 < \frac 12$ is NOT impossible. In fact it means $x$ MUST equal $y$. $\endgroup$ – fleablood Sep 14 '17 at 0:29
  • $\begingroup$ A radius of a set is a real number. It does not have to be a legitimate distance. Example a radius can be negative. If so, say $r = -3$ then the neighborhood is $\{x|d(x,y) < -3\} = \emptyset$. It's an empty neighborhood be it is well defined. Likewise let $X =\mathbb Q$. Let $r = \pi$. Then the neighborhood is $\{x\in \mathbb Q| d(x,y) < \pi\}$. That is perfectly okay, even though $d(x,y) = \pi$ is impossible. Likewise $\{x|d(x,y) < 1/2\}$ is well defined even if it is the discreet metric. $\{x|d(x,y)< 12\} = \{x\}$. $\endgroup$ – fleablood Sep 14 '17 at 0:45

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