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Let $X, Y, Z \neq \emptyset$. The set of elements that belong to exactly two of the sets $X,Y,Z$ was asked in the question and $(X\cup Y\cup Z) \setminus \left( \left(X\triangle Y\right) \triangle Z\right)$ was given as the answer I worked this out using Venn diagrams and found thisVenn diagram where unshaded portion gives the required value Here the unshaded portion gives the required value and we can see that this portion belongs to all 3 sets, which contradicts our assumption. I have just begun discrete maths, and I have been sitting with this for a long time. Could you please say where I am going wrong?

Asaf Karagila
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Ray
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  • "we can see that [the unshaded] portion belongs to all 3 sets". This is the part I'm not seeing. The part belonging to all three sets would be the dead center, which is shaded in your right hand sketch. – Teepeemm Feb 17 '22 at 21:41
  • Yes, you are right. I thought that the unshaded portions combined must belong to exactly two sets. Thank you. I should have read the question correctly. – Ray Feb 19 '22 at 10:53

2 Answers2

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This is the illustration of $X\Delta Y$:

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This is the illustration of $(X\Delta Y)\Delta Z$:

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Thus, convince yourself that $X\cup Y\cup Z \setminus ((X\Delta Y)\Delta Z)$ is the set of all elements that belong to exactly two of $X,Y,Z$.

Golden_Ratio
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$X \Delta Y \Delta Z$ is the set of all points that are in an odd number (i.e. 1 or 3) of the sets $X,Y,Z$. This is well-known and true for any number of "summands".

So $X \cup Y \cup Z \setminus (X \Delta Y \Delta Z)$ is the set that is in exactly two of them.

Henno Brandsma
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