Good day to everyone, can you please provide feedback on my personal attempt to prove this simple statement (a problem from Rudin's Principles of Math Analysis)?
Let A be the set of real numbers x such that $0 < \mathit{x} \leq 1$. For every $\mathit{x} \in \mathbf{A}$ let $\mathbf{E}_x$ be the set of real numbers y such that $0 < \mathit{y} < \mathit{x}$. Then the intersection $\bigcap_{x \in A} \mathbf{E}_{x}$ is empty.
My proof (by contradiction): Assume the intersection is not empty. That means at least one y exists which belongs to every $\mathbf{E}_{x}$. Since it belongs to every $\mathbf{E}_{x}$, obviously y has to be the smallest possible real number within A (if there was some x for which $y > x$, then y would not be a common element for every possible $\mathbf{E}_{x}$). 0 is the infimum of A. If y is the smallest element and $0 < y$, 0 is clearly not the infimum, which contradicts our definition and finishes the proof.
Please point out any flaws in my reasoning, I am trying to get back to understanding solid mathematics.
You've got the right idea. Maybe say that $y\le x,\,\forall x\in A$, instead of "$y$ is obviously the least element in A", since $A$ has no least element.
Direct proof:
Let $y\in A$. Then $\frac y2\in A$ and $0\lt \frac y2\lt y$. Thus $y\not\in E_{\frac y2}$. Thus $y\not\in\bigcap_{x\in A} E_x$. Thus $\bigcap_{x\in A} E_x=\emptyset$.