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I'm looking at Bungo's answer in: Proof that the set of irrational numbers is dense in reals

and in the last step it says: "Since $\mathbb{Q}+\sqrt{2}$ is a subset of the irrationals, we conclude that the irrationals are also dense in $\mathbb{R}$.

$Q1$: What result was used in this step? Is it "If a set $X$ contains a subset dense in $\mathbb{R}$, then $X$ is also dense in $\mathbb{R}$?" How would we show this?

$Q2$: Is $\mathbb{Q}+\sqrt{2}$ the same set as the field extension $\mathbb{Q}(\sqrt{2})$ (the field obtained by adjoining $\sqrt{2}$ to $\mathbb{Q}$)?

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    Here $\mathbb{Q}+\sqrt{2}$ means all reals of the form $r+\sqrt{2}$, where $t$ is rational. No significant connection to the field extension. – André Nicolas Jan 31 '16 at 02:49
  • @AndréNicolas where $r$ is rational? – user167857 Jan 31 '16 at 02:55
  • What is your definition of "$X$ is dense in $\mathbb R$"? Is it "for any $a,b\in\mathbb R$ with $a\lt b$ there is an $x\in X$ such that $a\lt x\lt b$"? Is it "Every nonempty open subset of $\mathbb R$ contains an element of $X$? Is it "the closure of $X$ is equal to $\mathbb R$"? – bof Jan 31 '16 at 02:56
  • @bof Hello bof, the first one is the definition I was given – user167857 Jan 31 '16 at 02:56
  • Yes, sorry, I was typing in the dark, and $r$ and $t$ are close on the keyboard. – André Nicolas Jan 31 '16 at 02:56
  • Q1: Suppose $X$ has a subset, call it $Y$, which is dense in $\mathbb R$. Consider any $a,b\in\mathbb R$ with $a\lt b.$ Since $Y$ is dense in $\mathbb R$ there exists $x\in Y$ such that $a\lt x\lt b$. Since $x\in Y$ and $Y$ is a subset of $X$ it follows that $x\in X.$ Thus we have found an element $x\in X$ such that $a\lt x\lt b.$ What part where you stuck on? – bof Jan 31 '16 at 03:00
  • @bof Not stuck anymore, thank you – user167857 Jan 31 '16 at 03:09
  • 1st def. If $X \subset Y$ and $X$ is dense. Dense means there exists an x in X with a certain property. That very same x is also a point of Y there is a point in Y with the same property (because it is the same point). So Y is dense. Same argument for Def 2 As for def 3, if X $\subset$ Y then the closure of X is a subset of Y (can't of a limit point of X that isn't a limit point of Y) and the closure of X is $\mathbb R$ so the closure of Y has the reals as a subset which means it is the reals. – fleablood Jan 31 '16 at 03:14

1 Answers1

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His logic is that $A= \mathbb Q + \sqrt 2 = \{q + \sqrt 2| q \in \mathbb Q\}$.

A is a set consisting entirely of irrationals. $A$ is dense in $\mathbb R$ because $\mathbb Q$ is and $A$ is really just a shift.

Q1) Yes if $X \subset Y \subset Z$ and $X$ dense in $Z$ then $Y$ is dense in $Z$. Every neighborhood in $Z$ contains a point of $X$ so every neighborhood in $Z$ contains a point of $Y$. So $Y$ is dense in $Z$.

Q2) No. It's important that $A$ is a subset of the irrationals so that it doesn't contain any rationals. Other wise it wouldn't prove anything.

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