I am trying to show that the image of an empty set is empty. Now I am aware of this post: What is an image of empty set?. But I am not entirely satisfied with the answer. I want to prove this without using contradiction. Here is my attempt at recreating the proof in the post:
Let $y$ be any element in an arbitrary ambient space. We wish to show that $y \not \in f(\emptyset)$. We know that if $x \in \emptyset$, then $y = f(x) \in f(\emptyset)$. However, this is vacuously true, hence $y \in f(\emptyset)$. But this is not true.
Now I understand that this argument is obviously false. But where is the logical error here?
Best Answer
It's less a logical error but more a wrong initial assumption.
By definition of the empty set there are no elements in it. Hence the statement $x\in\emptyset$ is always false, or in other words, an $x$ turning the statement to be true does not exist.
Your argumentation starts with "We know that if $x\in\emptyset$, then $\dots\,$", and this compound statement is true by the preceding. Which gives you a free ticket to conclude$-$or to declare$-$that any statement which is inserted for the dots is true.
In this context you may read the answer by Gadi A, and take account of Masacroso's comment hereof.