# Inclusion vs Belonging, and the proof of empty set being subset of every set

I just learnt the difference between Inclusion ($$\subset$$) and Belonging ($$\in$$). I also learnt as an example,

for the two sets $$A=\{4\}$$ and $$B=\{4, 5\}$$, $$A \not\subset B$$, but $$A \subset \{\{4\},5\}$$. -- (I)

However, Halmos's 'Naive Set Theory' book, as well as few posts like this and this say something like

"It ($$\varnothing \subset A$$) could be false only if $$\varnothing$$ had an element that did not belong to A."

Wondering if my understanding of (I) is wrong! Otherwise, I can see that even if A has no element that does not belong to B, A is not a subset of B.

• (I) is false. $A \subset B$ and $A \in \{\{4\},5\}$. By definition, for two sets $A,B$ we have $A \subseteq B$ precisely if every element that is contained in $A$ is also contained in $B$ – G. Chiusole Aug 20 '19 at 14:58
• Maybe have a look at this question – G. Chiusole Aug 20 '19 at 15:04

Your (I) is indeed false. "$$A\subseteq B$$" means "Every element of $$A$$ is an element of $$B$$," and it should be clear that every element of $$A$$ (there's only one - $$4$$) is also an element of $$B$$.
I've used "$$\subseteq$$" instead of "$$\subset$$" since some texts use the latter to refer to proper subsets. Also, it's worth noting that the inclusion/belonging terminology isn't universal mathematically, and in particular "$$A$$ is included in $$B$$" could reasonably be misunderstood as "$$A\in B$$."
Now if you replace your $$A$$ with "$$\{\{4\}\}$$," then the statement is indeed true, and it's possible (likely, even) that this is what was intended and that there was a typo or mishear/misreading issue.