suppose we replace "nearer than 10" with something a bit less ambiguous (we will assume our relation is on integers, just to be specific about which set we are relating):

a~b if (and only if) |a-b| < 10.

it is clear that this is reflexive:

a~a because |a-a| = |0| = 0 < 10.

it IS symmetric because if a~b, then |a-b|= |b-a| < 10, so b~a.

but what about transitivity: is it true that if:

a~b and b~c, that a~c?

no.

suppose a = 4, b= 13, c = 17.

then a~b because |a-b| = |4-13| = |-9| = 9 < 10.

and b~c because |b-c| = |13-17| = |-4| = 4 < 10.

however, we do NOT have a~c, because |a-c| = |4-17| = |-13| = 13 > 10.