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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?

Hanno
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Mathematics_Beginner
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    Vacuous truth means that "if P, then Q" is true because P is false, and this fact holds irrespective of the truth-value of $Q$. Thus, you cannot conclude that $Q$ is true (neither that it is false). – Mauro ALLEGRANZA Dec 11 '21 at 10:01
  • @MauroALLEGRANZA I think this is probably trivial, but I am kind of stuck at showing this. I agree what you are saying is true. So I instead wrote: we know that $y \in f(\emptyset)$ iff there exists some $x \in \emptyset$ such that $f(x) = y$. Therefore to show otherwise, we need to show for all $x \in \emptyset$, $f(x) \not= y$. It is not immediately obvious why this is true. Since again this would be vacuously true, and according to your comment, we can not conclude that $f(x) \not= y$? – Mathematics_Beginner Dec 11 '21 at 22:48
  • @MauroALLEGRANZA I might have see it now as I just started typing the comment. We simply need the IMPLICATION to be true in order to prove our statement. Therefore, it does not matter if $f(x) = y$ or not, the implication is always correct, hence $y \not \in f(\emptyset)$ for any $y$. Is this correct? – Mathematics_Beginner Dec 11 '21 at 22:51
  • $x\in \emptyset$ is not vacuously true. It is vacuously false. – MJD Jun 17 '22 at 09:32
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    @Mathematics_Beginner If you are happy with the given answer you may action to accept it. This also gives your question the "Answered" status. – Hanno Jul 02 '22 at 11:51

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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.

Hanno
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