Why is this set determined to be empty? From Eccles' Introduction to Mathematical Reasoning, problem 7.1 asks you to determine the set for:
$${\{n \in \mathbb{Z}^+ \mid \forall m \in \mathbb{Z}^+, m \leq n \}}$$
The answer provided in the back of the book is $\emptyset$. Why is $\{1\}$ not an answer? It satisfies $\mathbb{Z}^+$ and $m \leq n$, does it not?
 A: Is $1\geq m$ for every positive integer $m$? In words, the set is the set of all positive integers greater than or equal to all positive integers.
A: In your question, m is an arbitrary integer. You are going to pick an integer n which satisfying the condition for all m belongs to Z, m <=n.
If you are choosing n=1 then it fails for m=2.
So we come to know that there is no such integer n.
Thus the set having no elements.
Empty set.
A: $$\{n \in \mathbb{Z}^+ \mid \forall m \in \mathbb{Z}^+, m \leq n \}$$
To say that $1$ is a member of this set means that the statement $\forall m\in\mathbb Z^+, m \le n$ is true when $1$ is put in the role of $n$. The expression $n\in\mathbb Z^+$ identifies $n$ as the variables in whose place $1$ must appear. Thus you'd be saying
$$
\forall m \in \mathbb Z^+, m \le 1.
$$
Thus every member of $\mathbb Z^+$ would be $\le 1.$
A: Suppose that the statement below holds:
\begin{align}
&\exists x(x\in\Bbb Z^+\land\forall y(y\in\Bbb Z^+\to y\le x))\\
&\alpha\in\Bbb Z^+\land\forall y(y\in\Bbb Z^+\to y\le \alpha)&&\text{so there is an $\alpha$ satisfing the statement}\\
&\forall y(y\in\Bbb Z^+\to y\le \alpha) && \text{pick $y\in\Bbb Z^+$ such that $\alpha <y$}\\
&y\le\alpha\land\alpha<y\\
&y<y && \text{but '<' relation isn't reflexive, i.e. $\forall x\neg(a<a)$}\\
&\bot
\end{align}
It shows that our initial assumption was wrong and there is no such element exists.
