The Composition of Continuous Functions is Continuous

Manuel Kauers TSW-Homework

A function f is called continuous if

∀_{y} ∀_{ε>0} ∃_{δ>0} ∀_{x} |x − y| < δ → |f(x) − f(y)| < ε.

We claim that if f and g are continuous then f ∘ g: x ↦ f(g(x)) is continuous as well. Proof: Assume that f and g are continuous, i.e.

∀_{y} ∀_{ε>0} ∃_{δ>0} ∀_{x} |x − y| < δ → |f(x) − f(y)| < ε  (A 1)
∀_{y} ∀_{ε>0} ∃_{δ>0} ∀_{x} |x − y| < δ → |g(x) − g(y)| < ε  (A 2)

We want to show

∀_{y} ∀_{ε>0} ∃_{δ>0} ∀_{x} |x − y| < δ → |f(g(x)) − f(g(y))| < ε  (G 1)

For arbitrary but fixed y_{0}, ε_{0} with

ε_{0} > 0,  (A 3)

we have to show

∃_{δ>0} ∀_{x} |x − y_{0}| < δ → |f(g(x)) − f(g(y_{0}))| < ε_{0}.  (G 2)

In (A 1) we can take y = g(y_{0}) and ε = ε_{0} (using (A 3)) obtaining

∃_{δ>0} ∀_{x} |x − g(y_{0})| < δ → |f(x) − f(g(y_{0}))| < ε_{0}.  (A 4)

Now we take such a δ = δ_{0}, i.e. assume

δ_{0} > 0,  (A 5)
∀_{x} |x − g(y_{0})| < δ_{0} → |f(x) − f(g(y_{0}))| < ε_{0}.  (A 6)

For this δ_{0} and for y = y_{0} we know from (A 2) and (A 5) that

∃_{δ>0} ∀_{x} |x − y_{0}| < δ → |g(x) − g(y_{0})| < δ_{0}.  (A 7)

Fix such a δ and call it δ_{1}, then we know that

δ_{1} > 0,  (A 8)
∀_{x} |x − y_{0}| < δ_{1} → |g(x) − g(y_{0})| < δ_{0}.  (A 9)

We will show that δ_{1} manages to play the role of δ in (G 2) by showing

∀_{x} |x − y_{0}| < δ_{1} → |f(g(x)) − f(g(y_{0}))| < ε_{0}.  (G 3)

To show (G 3), we show

|x_{0} − y_{0}| < δ_{1} → |f(g(x_{0})) − f(g(y_{0}))| < ε_{0}  (G 4)

for an arbitrary but fixed x_{0}. Note that δ_{1} > 0 by (A 8). By (A 9) we know that

|x_{0} − y_{0}| < δ_{1} → |g(x_{0}) − g(y_{0})| < δ_{0}  (A 10)

and taking x = g(x_{0}) in (A 6) yields

|g(x_{0}) − g(y_{0})| < δ_{0} → |f(g(x_{0})) − f(g(y_{0}))| < ε_{0}.  (A 11)

Finally, by the transitivity of implication, (A 10) and (A 11) give

|x_{0} − y_{0}| < δ_{1} → |f(g(x_{0})) − f(g(y_{0}))| < ε_{0}  (A 12)

quod erat demonstrandum.