I just saw this and I thought I should emphasize it, because an examiner would rightly reject this definition. Saturation describes an equilibrium condition, viz. that the solution contains the same amount of solute that would be in equilibrium with undissolved solute.
For a salt (or solute), #MX#, we describe the equilibrium reaction:
#MX(s) stackrel("solvent"" ")rightleftharpoons M^+ + X^-#
Now there is a solubility product, #K_"sp"=[M^+][X^-]#, and if the ion product #[M^+][X^-]=K_"sp"#, then the solution is #"SATURATED"# with respect to #MX#.
If #[M^+][X^-]>K_"sp"# then the solution is #"SUPERSATURATED"# with respect to #MX#.
And the solution is #"UNSATURATED"# with respect to #MX# if #[M^+][X^-]# #<# #K_"sp"#. If you can remember how to frame these expressions, it will save you a lot of angst in an exam.
I reiterate that a statement to the effect that #"the solvent holds all the solute that it can"# to define #"saturation"# would be REJECTED OUTRIGHT by an examiner, in that a supersaturated solution manifestly contains more solute than a saturated one.
A #"supersaturated solution"# thus contains an amount of solute #"GREATER"# than would be in equilibrium with undissolved solute.
A supersaturated solution may be formed by heating the solution with excess solute, and cooling very carefully. This metastable solution may be brought back to saturation (to equilibrium) by introducing a seed crystal or scratching the container with resultant precipitation of a mass of crystals from solution. This is the basis for many spectacular classroom demonstrations of supersaturation (usually with sodium thiosulfate).
An #"unsaturated solution"# likewise contains less dissolved solute than would be in equilibrium with undissolved solute,
i.e. If #[M^+][X^(-)]# #<# #K_"sp"# then the solution is #"unsaturated"#. #"Saturation/unsaturation/supersaturation"# thus all refer to an equilibrium property