Problem with dense set

Question

On ' Set theory with an introduction to real point sets'(Dasgupta, Abhijit ,2014) i found this exercise:

This is interesting because compare the topological (left,1) and order (right,2) definition of dense set.

(1) $A$ is dense in $X$ if $X\subseteq \overline{A}$.

! Ps. I considered $X$ as subset of $\mathbb{R}$ and then $\overline{A}$ is closure of $A$ (i.e. set of adherent point of $A$)

(2) forall $x,y\in X$ with $x<y$ there is $a\in A$ with $x<a<y$.

Recall also that: (3) $X$ is dense order if forall $x,y\in X$ with $x<y$ there is $z\in X$ with $x<z<y$.

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My proof of 1)-->2): Let $x,y\in X$ with $x<y$.

For (3) exists $z\in X$ with $x<z<y$.

For (1) $z\in \overline{A}$. From definition of adherent point exists $c\in (x,y)\cap D$ and ok.

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I have problem with 2)-->1) implication. If i considered $X=[0,1)\cup\{2\}$ and $A=[0,1)$ then it seems to me that: $X$ is a dense order, $A$ verified (2) but not (1) ($\overline{A}=[0,1]$).

What am I doing wrong?

Answer 1

By definition $493$ of Dasgupta, $2$ is indeed a limit point of $A$ considered as a subset of the order $X$.

Does the so called topological definition appear in the book? I couldn't find it. What are you doing wrong? Well, you are using a definition that doesn't appear in the book and in fact contradicts the definition appearing in the book. Different textbooks might (and very often do) use different definitions or conventions.