Prove that there exists a unique set $T$ such that for every set $S$, $S\cup{T}=S$.

Question

Prove that there exists a unique set $T$ such that for every set $S$, $S\cup{T}=S$.

So far I have assumed that there exist two sets, $T1$ and $T2$ such that $S\cup{T1}=S$ and $S\cup{T2}=S$.

Not sure where to go about this now, any help would be appreciated thank you!

Answer 1

By taking $S=\varnothing $ we get $\varnothing\cup T=\varnothing $, hence $T=\varnothing $, thus proving uniqueness.

Answer 2

From your working we have $$S \,\,\cup\,\, T1=S \,\,\cup\,\, T2\,\,(=S)$$ $$(S \,\,\cup\,\, T1)\setminus S=(S \,\,\cup\,\, T2)\setminus S$$ $$\therefore T1=T2$$ So, we have that $T$ is a unique set. We can see that it must also be the null set - $\varnothing$ - as its union with the set $S$ does not add any elements to $S$.

Answer 3

$$T_1 \overset{(a)}= T_1 \cup T_2 = T_2 \cup T_1 \overset{(b)}= T_2$$ where at $(a)$, we used that $S = S \cup T_1$ for every set $S$, and at $(b)$, we used that $ S \cup T_2 =S $ for every set $S$.