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.

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    $\begingroup$ (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 $B$ $\endgroup$ – G. Chiusole Aug 20 '19 at 14:58
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    $\begingroup$ Maybe have a look at this question $\endgroup$ – 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.

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  • $\begingroup$ @ Noah Schweber - You are very generous with your last para, but it was more of wrong understanding from my part. Thanks to you and 'G. Chiusole'. $\endgroup$ – KGhatak Aug 20 '19 at 15:23

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