$\forall$ real $r \gt 0$, $\exists$ and natural number $M$ such that $\forall$ natural numbers $n>M$, $0 \lt \frac{1}{n} \lt r$
I think I understand this up until the last part $0 \lt \frac{1}{n} \lt r$. I believe this is true because from my perspective it seems like it is saying 'there is a number $n$ bigger than $M$ and if you take the reciprocal of that $n$ then it will be between $0$ and $r$
Is my assessment correct?
Let $x$ be a real number. If $7 \le x \lt 7+ \epsilon$ for all $\epsilon \gt 0$, then $x=7$
I believe this is true because my value of $x$ is either bigger than or equal to $x$, and since it smaller than $7+\epsilon$ that means it cannot take on values of $x$ in the neighborhood of $7$, so it must be $7$, right?
For any set $S$ of real numbers, every boundary point of $S$ is either an upper bound or a lower bound of the set $S$.
I feel like this somewhat confusing because of the infimum and supremum ideas. My gut tells me this false because does not have to be an upper or lower bound.
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$\begingroup$Your assessment of the first statement is correct. It's actually equivalent to saying that $\Bbb{N}$ is unbounded above. For suppose $\Bbb{N}$ were bounded above by some $x \in \Bbb{R}$; then let $n = \sup(\Bbb{N})$. Since for all $a, b \in \Bbb{N}$ we have $a = b$ or $|a-b|\geq 1$, it follows that $n \in \Bbb{N}$. But then by the induction property $n+1 \in \Bbb{N}$, and so $n$ is not an upper bound, a contradiction.
Your reasoning for the second problem is correct, though you could make it more rigorous.
The third statement is false, for let $S = [a, b] \cup [c, d]$ for $a < b < c < d$. $b$ is in the boundary of $S$, but it is neither an upper nor lower bound of $S$.
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