Find the cardinality of $\{\{\{1, 4\}, a, b, \{\{3, 4\}\}, \{\emptyset\}\}\}$?

Question

So I processed to solve this: set within a set $$ \Bigg\{\bigg\{\{1, 4\}, a, b, \big\{\{3, 4\}\big\}, \{\emptyset\}\bigg\}\Bigg\} $$ by doing the following, finding the cardinality of $$ \bigg\{\{1, 4\}, a, b, \big\{\{3, 4\}\big\}, \{\emptyset\}\bigg\} $$ which has 5 elements in the set and then afterwards finding the cardinality of $\{5\}$ which has 1 element. Therefore $$ \Bigg|\Bigg\{\bigg\{\{1, 4\}, a, b, \big\{\{3, 4\}\big\}, \{\emptyset\}\bigg\}\Bigg\}\Bigg| = 1. $$ Is it correct or I have made a mistake?

Answer 1

I am no set-theorist, but we can see quite clearly that your set has only $1$ element which is a set itself that has as you said $5$ elements. So yes, $1$ is the answer.

Answer 2

I have edited your question and used set braces ($\{$ and $\}$) of varying sizes to make it even more obvious that the set you are dealing with is of the form $$ \{A\} $$ where $A$ is some mathematical object (in this case a set). So this means your set has $1$ element, hence has cardinality $1$. It doesn't matter what $A$ is at this point. It could be a number, a matrix, a function, a topological space, a set containing 5 elements, a set containing 100 elements, a set containing infinitely elements, or a picture of my grandmother.

I also wanted to say something about your explanation (most of this is repeating other comments). So again, the set you're interested in is of the form $\{A\}$, where $$ A= \bigg\{\{1, 4\}, a, b, \big\{\{3, 4\}\big\}, \{\emptyset\}\bigg\} $$ You said you calculated the cardinality of $A$ and got $5$. This is correct. Then you said you calculated the cardinality of $\{5\}$ and got $1$. This is also correct. Then you concluded that the original set has cardinality $1$. Again, correct. So all of these statements are individually correct, but notice that when you string them together they don't really create a logical argument and some of the statements become superfluous. You could have also said "I calculate the cardinality of $A$ and get $5$, then I calculate the cardinality of $\{\pi^2+e\}$ and get $1$. So my the original set has cardinality $1$." Do you see what I mean? The point is that $A$ having cardinality $5$ has no bearing on the cardinality of $\{A\}$.