Showing $\inf A=0$, for $A=\bigcup_{n=1}^{\infty}\{\frac{1}{n}\}\subset(0,1]$

Question

I have this problem given to me from my intro. to analysis course I was hoping you guys could help me with. Obviously, there's something I'm not understanding, so I appreciate any hints/suggestions. Thanks in advance.

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I claim that $\inf A=0$, so I need to find $a\in A$ such that $a<0+\epsilon,\forall\epsilon>0$. I thought of $a=\frac{1}{\epsilon}\in A$, but $\frac{1}{\epsilon}<\epsilon$ doesn't hold for epsilon less than one. I also thought of of $\frac{\epsilon}{2}$, but $\frac{1}{n}=\frac{\epsilon}{2}\implies n=\frac{2}{\epsilon}\not\in [1,\infty)$.

Answer 1

What you need is $a=\frac{1}{n} < \varepsilon$, which is true if you choose $n > \frac{1}{\varepsilon}$.

Answer 2

Don't think of your $a$ as a variable. Instead think of it as a function:

You're looking for a function $a(\epsilon)$, so that $\frac{1}{a(\epsilon)} <\epsilon$ .

Now all that's left to do is manipulate the equation so that $a(\epsilon)$ is isolated on one side: $$\frac{1}{a(\epsilon)} <\epsilon \Leftrightarrow a(\epsilon) >\frac{1}{\epsilon}$$

And then choose your function $a(\epsilon)$ so that it fulfills the inequality.