Lower Bound for subsets of $\mathbb{N}$

Question

Let $P = \{A \in \mathbb{N} : A \neq \emptyset, |A|<\infty, |A| = 2k, k\in \mathbb{Z}^{+}\}$. Consider $X = \{ A \in P: 1 \in P \}$. Does $X$ have a lower bound in $(P,\subseteq)$.

We can see that $X\subseteq P$, which contains the element $1$ of $P$. By definition, an element $p\in P$ is a lower bound for $X$ iff $\forall x \in X$ we have $p\subseteq x$. Let $p = \{ 1 , a_2, a_3,\cdots,a_{2k}\}$ be the elements of $P$ which contain $1$ and are organized in ascending order, and let $x = \{ 1 , b_2, b_3,\cdots,b_{2m}\}$, where $m\in \mathbb{Z}^+$, and organized in ascending order. By comparing terms, it is not necessarily that $a_{i} = b_{i}$, then, there does not exist a $p\in P: p\subseteq x, \forall x\in X$, so $X$ does not have a lower bound.

I am right or wrong,

Thanks

Answer 1

Your conclusion is correct, and you have the right idea for the argument, but your presentation of it isn’t very clear. A clearer presentation would be something like this:

Suppose that $p=\{1,a_2,\ldots,a_{2k}\}$ for some $k$ is a lower bound for $X$. If $k>1$, then $\{1,a_2\}\in X$, and $p\nsubseteq\{1,a_2\}$, so we must have $k=1$. Now let $b\in\Bbb N\setminus\{a_2\}$; then $\{1,b\}\in X$, and $p\nsubseteq\{1,b\}$. Thus, $X$ has no lower bound in $P$.

An even simpler argument is to note that $\{1,2\}$ and $\{1,3\}$ both belong to $X$ and have no lower bound in $P$.