Does Injectivity Ensure f(Aᶜ) is a Subset of f(A)ᶜ?

[bedi]

Homework Statement

Let $f: X \rightarrow Y$ be a function. Suppose $A$ is a subset of $X$. Show that if $f$ is injective, then $f(A^{c})\subseteq f(A)^{c}$.

Homework Equations

The Attempt at a Solution

If $x \in A^{c}$, then there is a $y \in f(A^{c})$ such that $f(x)=y$. As $f$ is an injection, $y \notin f(A)$, hence $y \in f(A)^{c}$.

Is that alright?

 

[Mark44]

Homework Statement

Let $f: X \rightarrow Y$ be a function. Suppose $A$ is a subset of $X$. Show that if $f$ is injective, then $f(A^{c})\subseteq f(A)^{c}$.

Homework Equations

The Attempt at a Solution

If $x \in A^{c}$, then there is a $y \in f(A^{c})$ such that $f(x)=y$. As $f$ is an injection, $y \notin f(A)$, hence $y \in f(A)^{c}$.

Is that alright?

You're not using the correct definition of one-to-oneness. For an injection, given a y value, there is exactly one x value. The usual example for a function that isn't one-to-one is f(x) = x². Here, both 2 and -2 map to 4.

It seems to me that what you are doing is pairing one number in the domain (x) with two numbers in the codomain (y₁ and y₂), where y₁ $\in$ f(A) and y₂ $\in$ f(A^C). That isn't even a function, let alone an injective function.

What you need to show is that if y $\in$ f(A^C), then it follows that y $\in$ [f(A)]^C. I think that's what you mean by f(A)^C.