Water is always in equilibrium with H3O+ and OH- ions. The following reaction is what we refer to as the autoionization of water.
\[\rm{2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}\]
You can see in this reaction that one water molecule is essentially transferring a proton to another water molecule. Thus water is acting both as an acid and as a base. We call such a molecule amphiprotic (both proton donor and proton acceptor).
The extent of this reaction is very, very small since the standard free energy of the water is significantly lower than the standard free energy of the ions. Nonetheless, it is important since these ion species are so reactive.
The equilibrium constant expression for this reaction is
\[\rm{K_w = [H_3O^+][OH^-]}\]
where we have K with the subscript w for "water". The water does not appear in the equilibrium expression since it is a liquid and its activity is taken as one. Kw is also called the "ion product for water".
Kw is a very small number. At room temperature it is very close to being exactly 10-14.
Note: This is a point of great confusion since this makes for very nice round numbers for concentrations at room temperature. It is important to note that: one, it is by chance that it ends up as such a nice round number in the units we have chosen since Kw is temperature dependent (like any equilibrium constant). Thus it has a different value (that is not such a nice round number) at different temperatures.
This means the product of the H3O+ and OH- concentrations is a constant (much like solubility). So if one of the concentrations increases, the other must decrease as the equilibrium adjusts. For example, imagine that we increase the H3O+ concentration by the addition of a strong acid. This will push the equilibrium back to the reactant side to try to "reduce" the H3O+ concentration. This will mean that the OH- concentration will be reduced.