Is it possible to prove uniqueness without using proof by contradiction? Until now, I've been presented to several proofs of uniqueness:


*

*$\emptyset$

*$1$

*$0$


For set theory, fields, etc. And it seems that they all rely in proof by contradiction. At least, at the present moment, I've searched a bit for it and all the proofs seems to employ this technique. So is it possible to prove it without it?
 A: You're probably structuring uniqueness proofs like, "We assume there are two different things, we prove they are the same, a contradiction." Just delete the word "different". Now it's not by contradiction.
This is a common quasi-error where people start by assuming the opposite out of some sort of training or conditioning, then don't actually use the false assumption.
Per your comment, I really do see why this feels against the grain. You really shouldn't think of a proof that says, "Suppose there are two objects satisfying a certain property, then *insert math* it turns out they were the same object all along!" as a contradiction.
With that attitude you'll probably fall for bogus proofs that $0=1$, like: Let $a=b$. Then $a-a=b-a$ so $\frac{a-a}{b-a} = 1$. But $a-a=0$ so $0=1$. If you don't see the flaw, here's a hint: think about what $b-a$ is. This is the kind of mistake you will make when your intuition thinks of something called $a$ and something called $b$ as two different objects because they have different labels. If $x=y$ is that a contradiction? No, that sounds silly and it is.
On a tangent, have you thought of the proof that $0.999\ldots = 1$? It's an amazing problem and many, many people have the wrong intuition - that the numbers must be different - because they are written differently. It turns out the same number may have two different ways to write it down by decimal expansion. People are very uncomfortable with that. They've never thought of numbers as abstract entities existing independently of their decimal representations, and if the decimal representations - ink on paper - are different, then the numbers must be. But this is false. For a given real number $x$, there are always (I think) two decimal representations, one ending in $000$s and one ending in $999$s.
A: You can phrase a uniqueness proof so that it's not a proof by contradiction, but it retains the same character. For example, let's suppose you want to prove that the additive identity in a field $F$ is unique.
Proof by contradiction: Let $0 \in F$ be an additive identity, and suppose there exists another additive identity $0' \in F$, with $0' \not= 0$. Then: $0 = 0 + 0' = 0'$. This contradicts the assumption that $0' \not= 0$.
Other proof: Suppose $0' \in F$ has the property that $a + 0' = a$ for all $a \in F$. Then consider the case when $a = 0$; we see that $0 = 0 + 0' = 0'$. Therefore the only additive identity in the field is $0$.
Same idea, different words :)
A: There is often no need for contradiction; to say that there is a unique object $x$ satisfying some formula $\varphi(x)$ is to say that


*

*There exists $x$ satisfying $\varphi(x)$ — symbolically, this is $\exists x\, \varphi(x)$;

*If $x,y$ are such that $\varphi(x)$ and $\varphi(y)$ are true, then $x=y$ — symbolically, this is $\forall x \forall y (\varphi(x) \wedge \varphi(y) \to x=y)$.


So you can prove uniqueness by first supposing $x$ and $y$ are objects for which $\varphi(x)$ and $\varphi(y)$ are both true, and deriving $x=y$. You've probably done this a thousand times without realising. For example


*

*There is a unique empty set. To see this, suppose that $A$ and $B$ are empty sets. For any $x$, the statements $x \in A$ and $x \in B$ are both false, so that $x \in A \Leftrightarrow x \in B$ is true. By the axiom of extensionality, $A=B$.

*Every group (or even monoid) has a unique identity element. To see this, let $G$ be a group and suppose $u,v \in G$ satisfy $ug=g=gu$ and $vg=g=gv$ for all $g \in G$. Then $u=uv$ since $v$ is an identity element, and $uv=v$ since $u$ is an identity element, so $u=v$.

*Every time you prove a function is injective, you're proving a uniqueness result. To say a function $f : X \to Y$ is injective is to say that, for all $y \in \mathrm{im}(f)$, there exists a unique $x \in X$ such that $f(x)=y$. This is proved by showing that if $x,x' \in X$ with $f(x)=f(x')$ ($=y$), then $x=x'$.

A: To prove uniqueness we use the injective property. 
Suppose we want to prove that the solution to $2x + 1 = 0$ is unique. 
Let $f(x) = 2x + 1.$ Let $a, b \in \mathbb R$. Assume $f(a) = f(b).$ Then $2a + 1 = 2b + 1$ implying $a = b.$ So, $f$ is injective and the solution is unique.
