Graduate Dense set vs no isolating points

[SchroedingersLion]

TL;DR

Differences between two concepts.

Greetings,

could you commend or correct the following:

A dense subset $X$ of a set $Y$ is a set such that in each environment of $y\in Y$, there is at least one element $x\in X$. In other words, the elements of $Y$ can be approximated arbitrarily well by elements in $X$.

A set with no isolating points is a set such that in each environment of $a\in A$, there are other elements of A.

Are dense subsets automatically sets with no isolating points?
Simple example, $\mathbb{Q}$ is dense in $\mathbb{R}$, so in each environment of $x\in \mathbb{R}$ there is at least one element $\in \mathbb{Q}$. But take an $x\in \mathbb{Q}$. Are there other rationals in each environment of a rational?SL

 

[member 587159]

Yes there are other rationals in each neighborhood of a rational! Given a rational $q\in \mathbb{Q}$, simply consider $q+\frac{1}{n}$ where $n$ is a sufficiently large natural number (basically this is rewording that $q+\frac{1}{n}\to q$).

Your definition of dense set seems ok to me. This is equivalent with saying that the closure of $X$ is $Y$ or that $X$ intersects every open set of $Y$.

A dense set can evidently have isolated points. Consider the space $Y = [0,1]\cup \{2\}$ (with the subspace topology inherited from the usual topology). Then $X=Y$ is dense in itself (trivially) but $2$ is isolated in $X$ since $\{2\}$ is an open set.

 

[SchroedingersLion]

Thank you Math_QED,

so while $\mathbb{Q}$ is dense in $\mathbb{R}$ and has no isolating points, it is generally not true that dense subsets have no isolating points.

 

[member 587159]

Thank you Math_QED,

so while $\mathbb{Q}$ is dense in $\mathbb{R}$ and has no isolating points, it is generally not true that dense subsets have no isolating points.

Exactly.