• Let S be a set and ~ a relation between any two objects in the set:
• Reflexivity For every a in S, a ~ a
• Symmetry For any a, b in S, if a ~ b -> b ~ a
• Transitivity For any a, b, c in S, if a ~ b and b ~ c –> a ~ c

Symmetry: if it goes one way, it goes back

The core idea, in plain words

A relation $\sim$ on a set $S$ is symmetric when every connection runs both ways:

Note the shape: it is an if–then. It never forces anything to be related. It only says that whenever a link exists, its mirror image exists too.

An everyday analogy: “is married to” vs “loves”

If Ana is married to Ben, Ben is married to Ana. Symmetric. “Loves”, sadly, is not: half of all poetry is about $a \sim b$ without $b \sim a$. And “is the sister of” is not symmetric either: Ana can be Ben’s sister while Ben is Ana’s brother.

A small worked example

On $S = \{1, 2, 3\}$:

• $R_1 = \{(1,2), (2,1), (3,3)\}$: symmetric ✔. Every pair has its partner; $(3,3)$ is its own mirror.
• $R_2 = \{(1,2), (2,1), (2,3)\}$: not symmetric ✘: $(2,3)$ is there, $(3,2)$ is not.
• $\varnothing$ (no pairs at all): symmetric ✔, vacuously. There’s no link to break the rule.

On the integers: $a = b$ ✔, $a \equiv b \pmod n$ ✔ (if $n \mid a-b$ then $n \mid b-a$), $a \le b$ ✘ ($1 \le 2$ but not $2 \le 1$).

The key counter-example: symmetric but not transitive

Let $a \sim b$ mean “$|a - b| \le 1$” on the integers (“are within one step of each other”).

• Symmetric ✔: $|a-b| = |b-a|$.
• Reflexive ✔: $|a-a| = 0$.
• Transitive ✘: $1 \sim 2$ and $2 \sim 3$, but $|1-3| = 2$, so $1 \not\sim 3$.

So it is reflexive and symmetric but not an equivalence relation. Friendship works the same way: my friend’s friend need not be my friend. “Is close to” is the everyday trap: lots of small “close” steps add up to “far”.

And the other way round, symmetric and transitive but not reflexive: on $S = \{1, 2\}$ take $R = \{(1,1)\}$. Symmetric ✔, transitive ✔, but $(2,2)$ is missing, so not reflexive. Element 2 is related to nothing at all, which is exactly the loophole in the “symmetry + transitivity gives reflexivity” false proof.

A picture in words: two-way streets

In the arrow picture, symmetric means every arrow comes with a return arrow, so you can redraw them as plain lines with no arrowheads. In the matrix picture, the table is a mirror image of itself across the main diagonal.

graph LR
  1((1)) --- 2((2))
  2 --- 3((3))

(Here $1$–$2$ and $2$–$3$ are linked both ways, but there is no $1$–$3$ line: symmetric, not transitive.)

Common misconceptions

• “Antisymmetric is the opposite of symmetric.” Not quite. Antisymmetric means “$a \sim b$ and $b \sim a$ imply $a = b$”. Equality is both symmetric and antisymmetric; the relation $\{(1,2),(2,1),(2,3)\}$ is neither. The true “never both ways” notion is asymmetric (e.g. $<$).
• “Symmetric means everything is related.” No: the empty relation is symmetric.
• “Symmetric relations must be reflexive.” No: “is married to” is symmetric, and nobody is married to themselves.

How it connects

• Together with Reflexivity and Transitivity it makes an equivalence relation; swapped for antisymmetry it gives a partial order. See Set Relations.
• A symmetric relation is the same thing as an (undirected) graph: friendship networks, road maps, molecules.
• In Groups, “same coset” is symmetric precisely because every element of a subgroup has an inverse in it.

The truth revealed: symmetry is the rule of the two-way street, and it says nothing at all about how far you can travel along it.