Venn Diagram of $A$, $B$ and $C$, such that $A \subseteq B$, $C \not \subseteq B$, $A \cap C \neq \emptyset$

Question

I am really confused as to how to draw a Venn diagram to this particular question. Because although $C$ is not a subset of $B$ it says that $A$ intersection $C$ is present. That would be great if someone could help me with that please :)

The question is "Give a Venn Diagram of $A$, $B$ and $C$, such that $A \subseteq B$, $C \not \subseteq B$, $A \cap C \neq \emptyset$"

Answer 1

You can also construct simple sets satisfying those conditions "by hand"; here with simple I mean finite sets of natural numbers. Take for example $B = \{1,2,3\}$. Then you could take $A = \{1,2\}$, so that $A \subseteq B$. Since $C \not\subseteq B$ but $A \cap C \neq \varnothing$, you can take $C= \{1,4\}$, so that $4 \in C$ but $4 \notin B$ (and hence $C \not\subseteq B$) and $1 \in A$ and $1 \in C$ (so $1 \in A \cap C$ and hence $A \cap C \neq \varnothing$).

It's always a good practice to try tackling problems with small examples, so that you get an idea of what's the flavour of the problem in hand.

Answer 2

Let $A=\{1,2\}$, $B=\{1,2,3,4,5,6\}$ and $C=\{2,3,4,5,6,7\}$. Since all the elements of $A$ are also contained in $B$, it follows that $A\subseteq B$. Note that $7\in C$ but $7\notin B$, thus $C\nsubseteq B$. Finally, since $2\in A$ and $2\in C$, then $A\cap C\neq\varnothing$.

Answer 3

Place $A$'s circle in $B$'s and make $C$'s big enough to not be entirely in $B$'s, and to somewhat overlap $A$'s. One special case $A\subset B\subset C$ nests circles.

Answer 4

All the answers encourage the OP to draw the Venn diagram on their own by providing hints/comments and concrete examples. We do the same here but encourage the OP to graph the work on paper and then produce a generic Venn diagram.

Recall the definition/notation of/for an interval of real numbers.

Set the universal set $U = [0,3] \times [0,3]$.

Since $A \subseteq B$ and $1 \lt 2$ define

$\quad A = [0,1] \times [0,3]$

and

$\quad B = [0,2] \times [0,3]$

and ✅ for that one.

We want to find (with a 'picture' in our head) a $\gamma$ so that with the definition

$\quad C = [\gamma,3] \times [0,3]$

Both

$\quad A \cap C = \emptyset$

and

$\quad C \not \subseteq B$

are true.

OK, how about setting $\gamma = 2$?

Extra Credit: Show that any $1 \lt \gamma \lt 3$ also works.