Yes, it is.

Check: Let the set be C, so that C has just one element, say C={a}. It needs to be true for all points x, y in C and all t in [0,1] that tx+(1-t)y is also an element in C. But C has just one element, and the only possible choices for x and y are to let them both be a, so that x=y=a. Then tx+(1-t)y = ta+(1-t)a = ta + a - ta = a, which certainly is an element of C.