Prove $(\forall x, y \in E)(\exists ! z\in Z)$ s. t. $x + z = y$.

Question

I am currently studying Introduction to Hilbert Spaces with Applications, the third edition, by Debnath and Mikusinski. Chapter 1, exercise 1, is as follows:

1. Prove that for every $x, y \in E$ there exists a unique $z \in E$ such that $x + z = y$.

The definition of vector space is given as follows:

By a vector space we mean a nonempty set $E$ with two operations:

$(x, y) \mapsto x + y$ from $E \times E$ into $E$ called addition.

$(\lambda, x) \mapsto \lambda x$ from $\mathbb{F} \times E$ into $E$ multiplication by scalars,

such that the following conditions are satisfied for all $x, y, z \in E$ and $\alpha, \beta \in \mathbb{F}$:

(a) $x + y = y + x$;

(b) $(x + y) + z = x + (y + z)$

(c) For every $x, y \in E$ there exists a $z \in E$ such that $x + z = y$;

(d) $\alpha (\beta x) = (\alpha \beta) x$;

(e) $(\alpha + \beta)x = \alpha x + \beta x$;

(f) $\alpha(x + y) = \alpha x + \alpha y$

(g) $1x = x$

Elements of $E$ are called vectors. If $\mathbb{F} = \mathbb{R}$, then $E$ is called a real vector space, and if $\mathbb{F} = \mathbb{C}$, $E$ is called a complex vector space.

Chapter 1 already contains a proof that it seems would prove this (modified by me):

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Let $x, y \in E$. By (c), there exists a $w \in E$ such that $x + w = y$. If $x + z_x = x$ for some $z_x \in E$, then, by (a) and (b),

$$y + z_x = (x + w) + z_x = (x + z_x) + w = x + w = y.$$

This proves that $z$ exists for every $x, y \in E$.

Now let $z_1, z_2 \in E$ such that $x + z_1 = x$ and $y + z_2 = y$ for all $x, y \in E$.

Therefore, we have that

$$z_1 + z_2 = z_1$$

and

$$z_2 + z_1 = z_2$$

Therefore, we can conclude that $z_1 = z_2$.

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I would greatly appreciate it if people would please take the time to review this proof for correctness.

Answer 1

Existence is guaranteed by (c). You just need to prove uniqueness. Suppose there exists $z_1,z_2 \in E$ such that $$x+z_1=y, x+z_2=y.$$ Then $$x+z_1=x+z_2.\tag{1}$$ By (c), corresponding to $x \in E$ and $0 \in E$ (whose existence and uniqueness are proved here), there exists $x' \in E$ such that $x+x'=0.$ Therefore by $(1),$ $$x'+(x+z_1)=x'+(x+z_2)$$ and finally (b) implies $z_1=z_2.$

Your solution actually shows the uniqueness of additive identity.

Answer 2

Let $z, z^\prime, w \in E$.

From (c), it follows that there exists a $w \in E$ such that $z + w = z^\prime$.

Let $x + z = y = x + z^\prime$ and $z + w = z^\prime$.

Then

$$y = x + z^\prime = x + z + w = y + w.$$

Therefore, we must have that $w = 0$.

And so, we have that

$$z^\prime = z + w = z.$$