I am almost completely new to analysis. I am asked to show the accumulation points of $A = \{x\in\mathbb{Q}\enspace |\enspace x^2<2\}$. I have an intuition that the answer is $[\text{-}\sqrt{2},\sqrt{2}]$ because I can't imagine any $“$gaps$"$ in this set. But I know this is nowhere near close to rigorous, and it is probably wrong. How do I prove that the proposed set has all the accumulation points of $A$? Please keep in mind that my working knowledge is very limited; this is not even an analysis course! I am familiar with the definition of an accumulation point and the Bolzano-Weirstrass Theorem, and that is about it.
Finding Accumulation Points of Set
general-topologyreal-analysis
Best Answer
Take any point $x\in [-\sqrt{2},\sqrt{2}]$ and $\varepsilon >0$. Then, by the density of the rational numbers, we can find $y\in A$ such that $y\in ( [x-\varepsilon, x)\cup (x,x+\varepsilon])\cap \mathbb{Q}$ (recall the definition of acumulation point). So, $x$ is an accumulation point of $A$, i.e. $x\in A'$. If $x\not\in [-\sqrt{2},\sqrt{2}]$ then we cna find $\varepsilon>0$ such that $[x-\varepsilon, x+\varepsilon)\cap A=\emptyset$.
Therefore $A'=[-\sqrt{2},\sqrt{2}]$.