# Unfamiliar Set operation

Here's the set:

$$\{x:x<8\} \setminus \{5\}$$

What does the "\" operator mean?

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It usually represents the set theoretic different en.wikipedia.org/wiki/Setminus#Relative_complement – Apostolos May 4 '11 at 16:58

for "relative complement", "set-theoretic difference".

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Thanks for pointing me in the right direction. – Fortisimo May 4 '11 at 17:23

It means that you are looking at all values less than $8$ without considering $5$. What you have written is not completely correct. If you are talking of real numbers less than $8$ and not equal to $5$ you will write it as $$\{x \in \mathbb{R}:x<8 \} \backslash \{5\} = \{x \in \mathbb{R}:x<5 \} \cup \{x \in \mathbb{R}:5<x<8 \} = (-\infty,5) \cup (5,8)$$ or if you are talking of natural numbers less than $8$ and not equal to $5$ you will write it as $$\{x \in \mathbb{N}:x<8 \} \backslash \{5\} = \{0,1,2,3,4,6,7\}$$ or if you are talking of integers less than $8$ and not equal to $5$ you will write it as $$\{x \in \mathbb{Z}:x<8 \} \backslash \{5\} = \{x \in \mathbb{Z}:x<5 \} \cup \{6,7\}$$ It is incorrect to write $$\{x:x<8 \} \backslash \{5\}$$ without specifying where $x$ belongs

In general, $A \backslash B = A \cap B^c$

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Thanks for the detailed explanation. Unfortunately, user9325 got there first. – Fortisimo May 4 '11 at 17:26

Intuitively similar (but not really) to minus. So if $A=\{1,2,4\}$ and $B=A\setminus\{2\}$, then $B=\{1,4\}$.

in your case, the set is $\{x:x<8, x\neq5\}$.