Let $x,y\in \mathbb{R}$
Statement 1: For all $x$ there exists $y$ such that $x\lt y$
Statement 2: There exists $x$ such that for all $y$ we have $x\lt y$
Statement 1 is true (choose $y=x+1$), but Statement 2 is false, but can't we just make $x=y-1$? I found the $x$ such that for any $y$ we choose, $x$ is less than it, and that solution is $x=y-1$. What am I missing?
The issue with statement $2$ is that this $x$ needs to work for each and every single possible $y$.
In the first statement, we have "there exists $y$" after stating what $x$ is; the $y$ in that scenario thus can depend on $x$.
In the second statement, however, we don't have such a benefit: "there exists $x$" comes first. Whatever $x$ would work needs to satisfy $x<y$ for $y=1$, for $y=100$, for $y=-10^{10^{10}}$, and so on and so forth: it needs to work for every possible number.
But that would mean that $x<y$ for every real number $y$ ... which can't be satisfied by any real number.