Thread: check my proof f(phi) = phi

1. check my proof f(phi) = phi

Hi

I need to check my proof. Suppose that $\displaystyle \mathbi{f}:\mathbi{X}\rightarrow\mathbi{Y}$

I have to prove that $\displaystyle \mathbi{f}(\phi)=\phi$

now here's my proof. Let $\displaystyle \small y_1\in f(\phi)$ be arbitraty
$\displaystyle \exists\ x_1\in \phi \ni f(x_1)=y_1$
But since $\displaystyle \phi$ is an empty set, there is no element $\displaystyle x_1$ in it. So function operation $\displaystyle f(x_1)$ can't be performed. In other words, $\displaystyle y_1\in \phi$ . This is one way proof.
Is the proof ok so far ? I also need to prove in other direction, but I am confused about it.

thanks

2. I take it that you're using the letter phi to stand for the empty set.

As far as I can tell, you must have left out something in the statement of the problem.

The following is NOT a theorem (of, say, set theory), so I don't know why you would be asked to prove it [edit: '0' for the empty set]:

If f:X->Y, then f(0) = 0.

Counterexample:

Let X = {0}, let Y = {1}, let f = {<0 1>}.

Then, also, your "proof" makes no sense. There's no basis to assume there is an x1 such that x1 is in 0 and/or f(x1) in 0.

What book is this problem taken from?

3. The LaTeX symbol for the empty set is \emptyset.

What do you mean by $\displaystyle f(\emptyset)$? Is it the case that $\displaystyle \emptyset\in X$, so $\displaystyle f$ is applied to an element of its domain? Or is it a special case of $\displaystyle f(A)$ where A is a subset, not an element of X? This latter thing, also sometimes denoted by $\displaystyle f[A]$, is defined as $\displaystyle \{f(x)\mid x\in A\}$ and is a subset of Y. In this case, $\displaystyle f(\emptyset)=\emptyset$ by definition, because there is no $\displaystyle x$ in $\displaystyle \emptyset$ that you can apply $\displaystyle f$ to.

4. Hi

I have taken the problem from "A friendly introduction to analysis:single and multivariable " second edition by Witold Kosmala. And the statement is

Suppose that $\displaystyle f:X\rightarrow Y$ Prove that
$\displaystyle f(\emptyset)=\emptyset$ where $\displaystyle \emptyset$ is an empty set.

Since both $\displaystyle f(\emptyset)=\emptyset$ and $\displaystyle \emptyset$ are sets, I am using
standard approach used to prove the equality of the sets. I have downloaded the
errata from the author's website at
Code:
http://www.mathsci.appstate.edu/~wak/
thanks

5. Hi emakarov

I have stated the complete problem. that's all the author has given. this is the first chapter in the book and in this section, he is covering functions. even I am not sure what he is trying to say

newton

6. And that symbol is ambiguous in general mathematical writing unless the author specifies it means 'subset' or 'proper subset', since some authors use it to mean 'subset' and other authors use it to mean 'proper subset'.

7. There must be some context missing here.