Injective Function: Proving Correctness with Singlets ##S## and ##T## in ##X##

[pepos04]

Homework Statement

Hi. Given a function $f$ from $X$ to $Y$, an exercise asked me to establish whether, given two subsets of $X$ (for example $T$ and $S$), therefore $f (T \cap S) = f (T) \cap f (S)$, if and only if $f$ is injective.

Relevant Equations

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I operated by placing $S$ and $T$ to two singlets belonging to $X$ and therefore established that for $T, S \in X$, therefore $f (T) = f (S) \implies S = T$, consequentially: $$f (T \cap S ) = f (T \cap T) = f (T) \cap f (T) = f (T) \cap f (S)$$. I would like to know if my procedure is correct and if there are still any shortcomings. Thank you.

 

[Steve4Physics]

Homework Statement: Hi. Given a function $f$ from $X$ to $Y$, an exercise asked me to establish whether, given two subsets of $X$ (for example $T$ and $S$), therefore $f (T \cap S) = f (T) \cap f (S)$, if and only if $f$ is injective.

I’m no mathematician but am wondering if you have stated the question completely and correctly.

As I understand it, the argument of a function must be an element of a set. So writing expressions such as $f (T \cap S)$ or $f (S)$ seems wrong and ambiguous.

Also, if $S$ and $T$ are singleton sets, I would have thought this would need to be explicitly stated in the question. It doesn't seem like an assumption you should need to make yourself for the question to make sense.

 

[pasmith]

Homework Statement: Hi. Given a function $f$ from $X$ to $Y$, an exercise asked me to establish whether, given two subsets of $X$ (for example $T$ and $S$), therefore $f (T \cap S) = f (T) \cap f (S)$, if and only if $f$ is injective.
Relevant Equations: \

I operated by placing $S$ and $T$ to two singlets belonging to $X$ and therefore established that for $T, S \in X$, therefore $f (T) = f (S) \implies S = T$,

This is a statement that $f$ is injective. Is that the direction of the equivalence you are attempting to prove? If so, please state that expressly.

If $T$ is a singleton subset of $X$, then it is incorrect to say that $T \in X$. If $T = \{t\}$ then $t \in X$ and $T \subset X$. But you don't have to name singleton sets; instead name the elements inside them (ideally as $x$ and $y$) since the definition of injectivity of $f$ is stated in terms of individual elements of $X$.

If you are assuming that $f$ is injective, then your aim is to prove a statement about arbitrary subsets of $X$. Better to start with the assumption that they are indeed arbitrary, rather than singletons.

I would start by showing that if $f$ is injective, then for all $x \in X$ and all $T \subset X$ we have $f(x) \in f(T)$ if and only if $x \in T$.

 

[pasmith]

I’m no mathematician but am wondering if you have stated the question completely and correctly.

As I understand it, the argument of a function must be an element of a set. So writing expressions such as $f (T \cap S)$ or $f (S)$ seems wrong and ambiguous.

By definition, if $f: X \to Y$ then for $T \subset X$, $$f(T) \equiv \{ f(x) : x \in T \} \subset Y$$ and for $W \subset Y$, $$f^{-1}(W) \equiv \{x \in X: f(x) \in W \} \subset X$$ where the latter is well-defined even if $f$ is not itself invertible.

 

[Steve4Physics]

By definition, if $f: X \to Y$ then for $T \subset X$, $$f(T) \equiv \{ f(x) : x \in T \} \subset Y$$ and for $W \subset Y$, $$f^{-1}(W) \equiv \{x \in X: f(x) \in W \} \subset X$$ where the latter is well-defined even if $f$ is not itself invertible.

Thanks. I didn't know that.

 

[pepos04]

The text asks me: given a function $f$ from $X$ in $Y$, prove that for every pair $S, T$ of subsets of $X$, equality $f(S \cap T) = f(S) \cap f(T)$ holds if and only if the function $f$ is injective.

 

[pepos04]

I exploited the fact that only with the injectivity of the function could that equality be achieved

 

[FactChecker]

Homework Statement: Hi. Given a function $f$ from $X$ to $Y$, an exercise asked me to establish whether, given two subsets of $X$ (for example $T$ and $S$), therefore $f (T \cap S) = f (T) \cap f (S)$, if and only if $f$ is injective.
Relevant Equations: \

I operated by placing $S$ and $T$ to two singlets belonging to $X$ and therefore established that for $T, S \in X$, therefore $f (T) = f (S) \implies S = T$, consequentially: $$f (T \cap S ) = f (T \cap T) = f (T) \cap f (T) = f (T) \cap f (S)$$. I would like to know if my procedure is correct and if there are still any shortcomings. Thank you.

This needs a lot of work. A valid proof of this would have several parts and you should be clear in each part, what you are assuming and prove one part at a time.
When proving any "A if and only if B", there are at least two things to prove:
1) If A then B
2) If B then A.
When proving that two sets, C and D, are equal, there are at least two things to prove:
1) $x \in C, \implies x \in D$
2) $x \in D \implies x \in C$

So I would expect a proof with several sections. State clearly what is being assumed in each section of the proof and make it clear what the result is.

 

[Steve4Physics]

The text asks me: given a function $f$ from $X$ in $Y$, prove that for every pair $S, T$ of subsets of $X$, equality $f(S \cap T) = f(S) \cap f(T)$ holds if and only if the function $f$ is injective.

The key phrase is 'every pair $S, T$ of subsets'. You could do this...

1) Show that if $f$ is injective, then $f (T \cap S) = f (T) \cap f (S)$ for all possible pairs $S, T$.

2) Show that if $f$ is not injective, then at least one pair $S, T$ can be chosen such that $f (T \cap S) \ne f (T) \cap f (S)$. The 'trick' is to make a suitable choice for $S, T$.

 

[PeroK]

This might help the OP, without giving too much away in terms of the solution to this particular problem.

For any subsets, $T, S$, and any function $f$, we have:
$$f(T \cap S) \subseteq f(T) \cap f(S)$$Proof:
$$y \in f(T \cap S) \ \Rightarrow \ \exists x \in T \cap S: f(x) = y$$Now
$$x \in T \cap S \ \Rightarrow \ x \in T \ \text{and} \ x \in S$$$$\Rightarrow \ y \in f(T) \ \text{and} \ y \in f(S)$$$$\Rightarrow \ y \in f(T) \cap f(S)$$QED