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The problem I am working on is:

Use a Venn diagram to illustrate the relationship $A⊆B$ and $B⊆C$.

From my understanding, I should be drawing several Venn diagrams, corresponding to the different situations that are possible.

Since $A \subseteq B$ does not specify whether $A$ is a proper subset or not, we have two situations:

$(1)$: $A=B$

$(2)$: $A \subset B$

The same two situations apply to $B \subseteq C$


(This part is just a side note, but I'd appreciate a corroboration of my reasoning) Also, by the transitive law, since for x we find in A, it is true that we can find it in B ($A \subseteq B$); and since for every x in B, it is true that we can find it in C ($B \subseteq C$), we can say that $A \subseteq C$. This is reasonable, because all of A is contained in B, and all of B is contained in C, so all of A must be contained in C.


Now, the different situations we have are as follows:

$(1)$: $A=B$ and $B \subset C$

$(2)$: $A \subset C$ and $B \subset C$

$(3)$: $A=B$ and $B=C$

$(4)$: $A \subset C$ and $B=C$

So, I would have to make a Venn diagram for each situation?

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    $\begingroup$ I think you just have to draw three concentric circles. The cases where an inclusion is mutual (equality) are implicit I think. Also, your case #2 is wrong; nothing prevents A and B to be disjoint with your statements. $\endgroup$ – Jonathan H Feb 25 '13 at 15:34
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Here's the list I came up with, with the "$\subsetneq$" symbol meaning "is a proper subset of":

$(1): A = B\;$ and $\;B = C.\;$ (And hence $A = C$ by transitivity)

$(2): A = B\;$ and $\;B \subsetneq C.\;$ (And hence $A \subsetneq C$, necessarily).

$(3): A \subsetneq B\;$ and $\; B = C.\;$ (And hence, $A \subsetneq C$, necessarily).

$(4): A \subsetneq B\;$ and $\; B \subsetneq C.\;$ (And hence, $A \subsetneq C$, necessarily).

Yes, the Venn diagram for each of the four scenarios would necessarily be different, if we are distinguishing between equality of sets and "being a proper subset of" a set, so you can cover all possible relationships by drawing a Venn diagram for each case, separately. For example, case $(1)$ would be a single circle, labeled A = B = C. Case two would be a circle A = B, contained within the larger circle C. etc...

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  • $\begingroup$ So, in short, my analysis is correct? $\endgroup$ – Mack Feb 25 '13 at 16:09
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    $\begingroup$ Where you have $A \subset C$, it should be $A \subset B$, since $A\subset C$ will then follow. And your use of $\subseteq$ should be $\subset$ in your cases (1) and (2) ($B\subset C$), since you're considering equality separately. Does that make sense? $\endgroup$ – amWhy Feb 25 '13 at 16:13
  • $\begingroup$ Yes, thank you! $\endgroup$ – Mack Feb 25 '13 at 17:07
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You're way over thinking this, and I know this question is old, but for anybody else who sees this in the future, this is what the answer should look like. That's why the question says show it in a Venn Diagram.

enter image description here

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Isn't the proposed solution the euler diagram, while the venn diagram would rather look like that?

venn diagram

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