Redox: Oxidation and Reduction
Look this as:

$$\ce{2Ca + O2 -&gt; 2CaO}\qquad\text{(complete reaction)}$$ 

then we can this split into two sub-reactions:

$$\ce{Ca - 2 e- -&gt; \overset{+2}{Ca}}\tag{1}$$

Here "$+2$" means that $\ce{Ca}$ lost $2$ electrons (or two negative units) and therefore it was oxidized. Since it lost something negative it has a higher number of protons (positive particles) remained. In other words the atoms is now positively charged. The initial state was $0$ (the same numbers of protons and electrons, or positive and negative particles/units) and it was increased by loosing two electrons (the number increased).

$$\ce{O + 2 e- -&gt; \overset{-2}{O}}\tag{2}$$

The oxygen got two electrons or two negative particles. Now oxygen has two negative particles more than positive particles (protons) and therefore it is $-2$. In other words, the number (oxydation state) decreased ($0 \to -2$).