$\forall$ 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.
Best Answer
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 $S$.