Week 1

 Hello!

My math experience has, unfortunately, been largely negative. I have always been in the "slow" math classes, due to the more abstract concepts that make up the bulk of disciplines such as algebra or calculus being difficult for me to grasp as a visual learner. Equations with no real-world applications have always been difficult for me. I've found mathematics to be mostly a source of stress and irritation- however, in the right fields (such as geometry or trigonometry,) I do very well. I'm not the most gifted math student- I certainly won't be earning any accolades for my mathematics performance- but I do very well when the math I'm doing has a practical, tangible real-world application. 

In my own words, as I understand them:

A set is any well and objectively defined group of elements that share something in common. For example, {The top 5 coolest dinosaurs} would not be a valid set, because "coolest" is a subjective term that cannot be clearly defined; however, {maniraptora} would be a valid set, because maniraptora is an objectively definable clade of dinosaurs. 

A subset is any and all sets of which all the elements are included in another set. For example, subsets of {maniraptora} could include {maniraptora}, {paraves}, {avialae}, {paraves, eumaniraptora}, etc. A proper subset, in comparison, is any subset of a set that is not the same as the given set. Proper subsets of {maniraptora} would be {avialae}, {eumaniraptora}, {troodontidae}, etc., but not {maniraptora} itself, as they are the same set. 

My definition of why the null set is a subset of all sets, copy-pasted from a late-night rambling while trying to figure the concept out:
So like. Its true by contraposition ("if A then B, so if not-A then also not-B.") So if X is not an element of set A, it must also not be an element of the null set- and we KNOW its not an element of the null set since the null set has no elements. So since its not-A and not-B, the vice versa must also be true (that X actually IS an element of the null set and so its also an element of set A.)