Open and Closed Sets - Sohrab Exercise 2.4.4 - Part 3

[Math Amateur]

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with a part of Exercise 2.2.4 Part (3) ... ...

Exercise 2.2.4 Part (3) reads as follows:

View attachment 7185I am unable to make a meaningful start on this problem ... can someone help me with the exercise ...

Note: Reflecting in general terms, I suspect the proof that $$\mathbb{N}$$ and $$\mathbb{Z}$$ are closed is approached by looking at the complement sets of $$\mathbb{N}$$ and $$\mathbb{Z}$$ ... visually $$\mathbb{R}$$ \ $$\mathbb{N}$$ and $$\mathbb{R}$$ \ $$\mathbb{Z}$$ and proving that these sets are open ... which intuitively they seem to be ... but I cannot see how to technically write the proof in terms of open sets and $$\epsilon$$-neighborhoods ... can someone please help ...

I have not made any progress regarding the set $$\{ \frac{1}{n} \ : \ n \in \mathbb{N} \}$$ ...

Peter
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The above exercise relies on the definition of open sets and related concepts and so to provide readers with a knowledge of Sohrab's definitions and notation I am provided the following text ...View attachment 7186

 

[Opalg]

The set $S = \{1/n:n\in\Bbb{N}\}$ is not open, because for example it contains the point $1$ but it does not contain any $\varepsilon$-neighbourhood of $1$. (In fact, the point $1 + \frac12\varepsilon$ is in that neighbourhood but is not in $S$.)

To show from the definition that $S$ is not closed, you are right that one must show that its complement is not open (because that is how closedness is defined). In this case, the point $0$ is in $S^c$, but every $\varepsilon$-neighbourhood of $0$ contains an element of $S$. In fact, given $\varepsilon>0$, choose (by the Archimedean property of $\Bbb{R}$!) an integer $n>1/\varepsilon$. Then $1/n \in B_\varepsilon(0)$.

 

[Math Amateur]

The set $S = \{1/n:n\in\Bbb{N}\}$ is not open, because for example it contains the point $1$ but it does not contain any $\varepsilon$-neighbourhood of $1$. (In fact, the point $1 + \frac12\varepsilon$ is in that neighbourhood but is not in $S$.)

To show from the definition that $S$ is not closed, you are right that one must show that its complement is not open (because that is how closedness is defined). In this case, the point $0$ is in $S^c$, but every $\varepsilon$-neighbourhood of $0$ contains an element of $S$. In fact, given $\varepsilon>0$, choose (by the Archimedean property of $\Bbb{R}$!) an integer $n>1/\varepsilon$. Then $1/n \in B_\varepsilon(0)$.

Thanks Opalg ... appreciate the help ...

Peter