If A is dense in (0, 1), then A is dense in R

Question

Is it true or false? Explain please. Thanks.

Dense definition: Let $∅ ≠ A, B ⊆ R$. We say that $A$ is dense in $B$, if $A ⊆ B$, and, in addition, if for every $x, y ∈ B$ such that $x < y$ there exists an $a ∈ A$ such that $x < a < y$

Answer 1

It is true. Let $A$ be dense in $(0,1)$. We have to show that for any two points $x,y \in [0,1]$ such that $x < y$ there is a point $a \in A$ such that $x < a < y$.

Notice that

$$0 \le x < \frac{2x+y}3 < \frac{x+2y}3 < y \le 1$$

and that $\frac{2x+y}3, \frac{x+2y}3 \in (0,1)$ since $x,y$ cannot both be $0$ or both be $1$.

Now since $A$ is dense in $(0,1)$, there exists $a \in A$ such that

$$\frac{2x+y}3 < a < \frac{x+2y}3 $$

so in particular $x < a < y$.

Answer 2

Yes, $A$ is dense in $[0,1]$.

Suppose you are given $x,y\in[0,1]$ with $x<y$. You want to find $a\in A$ such that $x<a<y$. You can't immediately use the denseness of $A$ in $(0,1)$, because it may be that $x=0$ or $y=1$. But you just have to adjust $x$ and $y$ a little, so that they do fall in $(0,1)$: put $$x'=x+\frac13(y-x)$$ $$y'=x+\frac23(y-x)$$ Now $x'$ and $y'$ are in $(0,1)$, so from the denseness of $A$ in $(0,1)$, there exists $a\in A$ such that $x'<a<y'$. And therefore $x<a<y$, which is what you want.

Answer 3

Consider an arbitrary ordered set $(A, R)$ and define the following notion of relative denseness as a binary relation on subsets of $A$: $$X \leqslant_{\mathrm{d}} Y \colon \equiv X \subseteq Y \wedge (\forall x, y)\left(x, y \in Y \wedge x<_R y \Rightarrow (x, y)_R \cap X \neq \varnothing\right)$$ Let us show that this binary relation is transitive, in other words given subsets $X$, $Y$ and $Z$ such that $X \leqslant_{\mathrm{d}} Y$ and $Y \leqslant_{\mathrm{d}} Z$ it follows that $X \leqslant_{\mathrm{d}} Z$. Indeed, by definition we have that $X \subseteq Y$ and $Y \subseteq Z$ so $X \subseteq Z$ follows by virtue of the transitivity of inclusion.

Let us now consider $u, v \in Z$ such that $u<_Rv$. As $Y$ is dense in $Z$, there exists $x \in Y \cap (u, v)_R$. Since $Y \subseteq Z$ we have in particular that $x \in Z$. Since $x, v \in Z$ and $x<_R v$, by the same token of $Y$ being dense in $Z$ we infer the existence of $y \in Y \cap (x, v)_R$. Since $x, y \in Y$ and $x<_R y$, we appeal now to the fact that $X$ is dense in $Y$ to infer that $X \cap (x, y)_R \neq \varnothing$. Let us note that by construction $u<_R x<_R y<_R v$, hence $(x, y)_R \subset (u, v)_R$ and subsequently $X \cap (u, v)_R \supseteq X \cap (x, y)_R \supset \varnothing$. This shows that $X$ is dense in $Z$.

In your particular case, the ambient ordered set is $(\mathbb{R}, \mathrm{O})$ (I have taken the liberty to denote the standard order on the reals by $\mathrm{O}$) and you are given that $A \leqslant_{\mathrm{d}} (0, 1)$. The fact that $(0, 1) \leqslant_{\mathrm{d}} [0, 1]$ is immediate so the above property of transitivity applies.

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Remark: given an arbitrary topological space $(X, \mathscr{T})$, one can define a similar binary relation of topological denseness between the subsets of $X$ by: $$M \subseteq_{\mathrm{d}} N \colon \equiv M \subseteq N \subseteq \overline{M},$$ which is equivalent to saying that $M$ is a dense subset of $N$ with respect to the subspace topology (on $N$, of course). This relation is also transitive, and in the particular case of a topological space induced by a total order, the two notions of relative denseness between subsets are (almost) equivalent.