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### Another rant on undergraduate Algebra

Far too much of Algebra is infested with “equivalence relation” and “well-defined mappings” and what not. Over time, I had grasped their essence, but faltered when trying to fathom their motivation/put them in context. I’m sure you’ve felt the same way in undergraduate Algebra.

Then I made the best decision of my life: I started reading “Elements of Set Theory” by Enderton. I think it is almost criminal if (Indian) colleges let students move on to Algebra and the like without first making them read Set Theory (especially from this book). But more on that later.

1. Equivalence relations- Why do we need equivalence relations? Intuitively speaking, we need them to clump elements together. Elements that we *want* to consider as the same. For example, if I wanted to clump all even numbers together as one element and all odd numbers as another, I would impose the condition “$p\sim q$ iff $2|p+q$ on $\Bbb{N}$, the set of natural numbers.

But do we really need equivalence relations to clump elements together which we consider to be the same? No! We can clump elements in whatever fashion we like, and not give a hoot about any specified relations. Heck, we do not even have to ensure that the clumps are disjoint! We can pretty much do anything we want.

However, if we have an equivalence relation, often things become “neat”. Resulting clumps become disjoint, etc. Hence, equivalence relations provide a useful tool, but by no mean the only tools to clump elements together.

2. Well-defined mappings- Often, to check whether a mapping (function, we say in our heads) is well-defined, we check whether the following condition is satisfied: $a=b\implies f(a)=f(b)$. Whut?! This is obvious! Why are we doing all this at all? It was not until recently that I realised the true value of this: this concept is tremendously useful when dealing with equivalence classes (or other kinds of clumps). Say you have $a\sim b$, where $a,b\in P$. Also, assume $f:P\to Q$, where $Q$ is any other set. Then if $f$ satisfies $p\sim q\implies f(p)=f(q)$, we can easily construct a new mapping $f^*$ from $f$, with $f^*:(P/\sim)\to Q$.

Essentially, the seemingly useless concept of “well-definedness” starts becoming useful when we have clumps (a more generalised form of equivalence classes).

### A rant on the null-set.

This as a post was long due.

What exactly is $\emptyset$? It is the single most confusing thing I have come across. I am going to try and elaborate its properties. I’m going to build the theory in stages.

1. Nothingness. Let us give it a symbol- $\mathfrak{not}$. This $\mathfrak{not}$ is an element of EVERY set. $\mathfrak{not}$ is the absence of elements, but it is present in everything. This is a little non-intuitive, if one were to take the physical world as a reference point. For $a$ to be present inside $S$, there has to be some trance, some sign of the presence of $a$ in $S$.

Maybe even a spaceĀ  ” “, showing an absence of elements. But really in set theory, we don’t have spaces of this kind to show an absence of elements. We have no such trace, which is fairly irritating. But one has to get used to it.

2. Now we come to $\emptyset$. This is NOT equivalent to $\mathfrak{not}$. $\emptyset$ is not the absence of elements. It is a set containing no element, the operating word being “set”. Technically, $\mathfrak{not}\in\emptyset$, which makes $\emptyset$ non-empty. However, we work around this by translating to English. When we say $\mathfrak{not}\in\emptyset$, what we’re really saying is no element belongs to $\emptyset$, thereby making it an emptyset. See? Clever!

3. Say we have a set $A=\{a,b,c\}$. Actually, $A=\{\mathfrak{not},a,b,c\}$. Hence, one element of $\mathcal{P}(A)$ is $\{\mathfrak{not}\}=\emptyset$, as stated before (by clever linguistics). Hence, $\emptyset\in\mathcal{P}(A)$ for every set $A$. In fact $\mathcal{P}(\emptyset)=\mathcal{P}(\{\mathfrak{not}\})=\{\{\mathfrak{not}\}\}=\{\emptyset\}$. Hence, $\emptyset\in\mathcal{P}(\emptyset)$ also.

4. Say we have to construct the set $S=\{x\in\Bbb{R}|x=x+1\}$. This set contains no element from $\mathbb{R}$. However, even THIS set has to contain the element $\mathfrak{not}$. Hence, $S=\{\mathfrak{not}\}=\emptyset$.

I have sort of developed my own concepts here, and have made set theory more understandable for myself. I hope it helps other readers too.