1

Given: $$H = \lambda f \in \mathbb{R} \rightarrow P(\mathbb{R}).\left\{ {x \in \mathbb{R}|x \notin f(x)} \right\}$$.

Show that $f(x) \ne H(f)$, for all $x, f$.

Well, this is my answer:
Let $y \in f(x)$.

By definition of $H$, $y \notin H(f)$.

Therefore, $H(f) \ne f(x)$.

is that good enough as a proof?

AnnieOK
  • 2,920

1 Answers1

1

No. (1) At "let $y \in f(x)$", you also have to consider the case $f(x) = \emptyset$. (2) The statement $y \not \in H(f)$ does not follow from the definition of $H$: $H(f) = \{x \in {\mathbb R} \mid x \not\in f(x) \}$, so $y \not\in H(f)$ is equivalent to $y \in f(y)$.

You'd better split cases: $x \in f(x)$ or $x \not \in f(x)$.

Magdiragdag
  • 15,049
  • What does it mean $x \in f(x)$? Maybe I get it wrong. Is it $x \in Im(f)$? – AnnieOK Mar 07 '14 at 13:33
  • Well, it's your question :-) The $f$'s under consideration are functions from $\mathbb R$ to $\wp(\mathbb R)$. So $f(x)$ is a set of real numbers and $x$, being a real number, either belongs to it or not. – Magdiragdag Mar 07 '14 at 13:37
  • Of course, but for my understanding $f(x)=y$ where $y$ is a real number, and not a set. That said, how can you define $x \in f(x) = y \in \mathbb{R}$? – AnnieOK Mar 07 '14 at 13:41
  • 1
    Well, your $f$'s don't have ${\mathbb R}$ as codomain, but $\wp({\mathbb R})$. So $f(x)$ simply is not a real number. If it were, the statement "$x \in f(x)$" would be meaningless. – Magdiragdag Mar 07 '14 at 13:46