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I came across the following question:

Given are the following relations on the functions from $\{1,...,10\}$ to $\{1,...,10\}$
R1 = {(f,g) | f(1)=g(1)}
Is this relation reflexive?

I don't know how to read the part, $f(1)=g(1)$ of this set builder notation. Are $f(1)$ and $g(1)$ referring to the first element of $\{1,...,10\}$?

2 Answers 2

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Let us denote $S = \{1,2, \ldots, 10\}$. Consider two functions $f, g: S \to S$, then we can ask if $f(1) = g(1)$ (that is: 'is the image of 1 under $f$ and $g$ the same?). If so, we define these functions $f$ and $g$ to be related and the relation is the set $R_1$. So for all functions $f,g$ such that $f(1) = g(1)$, we have that $(f,g) \in R_1$.

$\textbf{SPOILER:}$ Now this relation is

reflexive

since we have that

f(1) = f(1), $\forall f: S \to S$.

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    Is this reasoning correct? The function f and g can have all values of {1,....,10} as arguments, and R1 only contains (1,1). Because 1 maps to itself, this relation is reflexive. – 2017-02-26
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    Not really: we consider all functions, but are only interested in the image of 1 under all these possible functions. Consider the following three functions: $f, g, h$ with $f(1) = 3, g(1) = 3, h(1) = 7$, then $(f,g) \in R_1$ (since they map 1 onto the same element) but $(f,h) \notin R_1$ since they map 1 to different elements. Now clearly for all functions $t$ we have that $t$ and $t$ map 1 onto the same element! – 2017-02-26
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    The examples made it clear for me! – 2017-02-26
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They are saying that the pair $(f,g)$ is in the set $R_1$ if and only if $f(1)=g(1)$. The condition for being a pair in $R_1$ is equality for an argument of $1$. Does that make sense?

Can you now tell whether or not the relation is reflexive?