Stochastic_Jimmy
New member
- Joined
- Jan 10, 2013
- Messages
- 27
Hi, I was wondering what you all thought about the answer given to the following question.
This is how the question is phrased:
Given \(\displaystyle f : \mathbb{R} \to \mathbb{R} \) where \(\displaystyle f(x) = \lfloor x \rfloor \), the greatest integer in \(\displaystyle x \), find \(\displaystyle f^{-1}(A) \) for \(\displaystyle A = \{0,1 \} \).
And the given answer is \(\displaystyle f^{-1}(A) = [0,2) \).
I was wondering why the answer is not \(\displaystyle \: f^{-1}(A) = (-\infty,2) \).
My reasoning is that for \(\displaystyle A = \{0,1 \} \), we have \(\displaystyle f^{-1}(A) = \big\{x \in \mathbb{R} \,\big|\, f(x) \in \{0,1\} \big\} \). And if \(\displaystyle f(x) \) is the greatest integer in \(\displaystyle x \), then wouldn't all negative real numbers also be included in the preimage? That is, wouldn't \(\displaystyle f\big( (-\infty, 2) \big) \) also be 1, which is in \(\displaystyle \{0,1\}\)? Or am I misunderstanding something?
Thanks in advance for your comments. Much appreciated!
This is how the question is phrased:
Given \(\displaystyle f : \mathbb{R} \to \mathbb{R} \) where \(\displaystyle f(x) = \lfloor x \rfloor \), the greatest integer in \(\displaystyle x \), find \(\displaystyle f^{-1}(A) \) for \(\displaystyle A = \{0,1 \} \).
And the given answer is \(\displaystyle f^{-1}(A) = [0,2) \).
I was wondering why the answer is not \(\displaystyle \: f^{-1}(A) = (-\infty,2) \).
My reasoning is that for \(\displaystyle A = \{0,1 \} \), we have \(\displaystyle f^{-1}(A) = \big\{x \in \mathbb{R} \,\big|\, f(x) \in \{0,1\} \big\} \). And if \(\displaystyle f(x) \) is the greatest integer in \(\displaystyle x \), then wouldn't all negative real numbers also be included in the preimage? That is, wouldn't \(\displaystyle f\big( (-\infty, 2) \big) \) also be 1, which is in \(\displaystyle \{0,1\}\)? Or am I misunderstanding something?
Thanks in advance for your comments. Much appreciated!