Show that the inclusion map i:Q -> R defined by i(q)=q for all q in Q, is continuous where both Q (rational numbers) and R(real numbers) are given the order topology.
This provides an example of proving the continuity of an inclusion map.
Real Analysis - Let f be a function defined on all of R that satisfies the additive condition f(x+y)=f(x)+f(y) for all x,y belong to R
a- show that f(0)=0 and that f(-x)=-f(x) for all x belong to R.
b- show that ...
Real Analysis - a- Show that if a function is continuous on all of R and equal to 0 at every rational point then it must be identically 0 on all of R
b- if f and g are continuous on all of R and f(r)=g(r) at every r ...
Show that a set of real rational functions is a field. - NOTE: In this description, R represents the symbol for the set of real numbers. I couldn't find a way to type or copy the correct R symbol for the set of real numbers. Also, the parentheses in R(x) i ...