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I'd appreciate some guidance regarding the following 2 questions. Number 1 should be clear, and number 2 is more of a discussion:

  1. Let $X$ be a topological space. Let $E$ be a dense subset. Can $E$ be finite without $X$ being finite? Or countable?

  2. Let $X$ be a topological space. I came across the following definition of "isolated point": a point $x$ which is not a limit point of $X\setminus\{x\}$. Or, for a metric space: a point $x$ such that there is an open ball centered on $x$ not containing any other point of $X$. These definitions make no sense to me: how can a ball of a topological space $X$ contain points not from $X$? It sounds like there is some kind of bigger topological space containing $X$ which is implicitly referred to. In that context, what meaning can one give to "an isolated point of $X$"?

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What's "trivial" about this questions? –  Pedro Tamaroff Feb 23 '13 at 17:20
    
For 1), notice that if $X$ is Kolmogoroff, $E$ finite implies $X$ finite. –  Seirios Feb 23 '13 at 19:03

6 Answers 6

up vote 1 down vote accepted

(1) Let $X$ be the real line, topologized so that a subset $U$ is open if and only if either (i) $U=\emptyset$ or (ii) $0\in U$. The singleton $\{0\}$ is dense in $X$, even though $X$ is far from finite.

(2) We can also say that $x$ is an isolated point of $X$ if and only if $\{x\}$ is open in $X$. The bit about the ball could be better stated to say that "there is an open ball centered on $x$ that has only $x$ as an element"--that is, $\{x\}$ is an open ball. For an example of how this can happen in a metric space, let $X$ be any non-empty set you like, and define $d:X\times X\to\Bbb R$ by $$d(x,y)=\begin{cases}0 & x=y\\ 1 & \text{otherwise.}\end{cases}$$ This is a metric on $X$ called the discrete metric. It has the property that for any $x\in X$ and any real $0<r<1$, the ball centered at $x$ of radius $r$ is simply the singleton $\{x\}$--so all points of $X$ are isolated when considered in the topology induced by this metric.

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  1. Yes. If $X$ has the indiscrete topology (only $\emptyset$ and $X$ are open) then any nonempty subset is dense.

  2. It is not said that the ball contains points not from $X$. It merely does not contain any points from $X$ apart from $x$. If $X=\{\text{foo},\text{bar}\}$ with metric given by $d(\text{foo},\text{bar})=1$, then the $\frac12$ ball around foo contains no other points, thus making foo isolated.

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For your second question, it does not imply that the ball could contain things other than elements of $X$; it just means that $x$ is an isolated point of $X$ when there is an open ball around $x$ such that the only element of the ball is $x$ itself.

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  1. Let $X = \mathbb{R}$ and define the topology $\mathcal{T} = \{\mathbb{R}, \emptyset, \{0\}\}$. The set $\{1\}$ is dense in $(\mathbb{R}, \mathcal{T})$ (it's closure is clearly all of $\mathbb{R}$, and it's a finite set.

  2. Consider $X = [0,1] \cup \{2\}$ with the usual Euclidean metric. The ball of radius $1$ about the point $2$ (in $X$) is just the set $\{2\}$. You don't need to have a larger space in mind when you talk about the isolated points of $X$. The ambient space $\mathbb{R}$ doesn't play any role here.

Alternatively, you could think about the metric $d(x,y) = 1$ if $x \neq y$ and $0$ otherwise on $\mathbb{R}$. Then the ball of radius $1$ about any point in $\mathbb{R}$ is the singleton $\{x\}$.

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1 - It dependes on the topology of the space $X$. For example, if you assume that the only open sets are $\emptyset$ and $X$, then, every set is dense in $X$.

2- In your definition it is said that there is an open ball, such that the only point of $X$ in this ball is $x$. Note that it is not saying that there is point outside $X$ in it.

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In the particular case where $X$ is a metric space, we know that a point in $X$ is a closed subset so a finite subset is closed as being finite union of closed subset and then it's itself closure , hence a finite subset can not being dense in $X$ if it's not finite.

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