Statement
The definitions of continuity for a function f:(X,d)→(Y,d′)
from one metric space to another is the same as f:(X,J)→(Y,K)
being continuous (where the topologies are those induced by the metric are the same, that is
- ∀a∈X∀ϵ>0∃δ>0:x∈Bδ(a)⟹f(x)∈Bϵ(f(a))
- ∀V∈K:f−1(V)∈J
Proof
⟹
Suppose f:(X,J)→(Y,K)
is continuous.
Let V∈K
- that is V
is open within Y
Let x∈f−1(V)
be given.
Then because V
is open, ∃ϵ>0
such that Bϵ(f(x))⊂V
(note that f(x)∈V
by definition of where we choose x from).
But by continuity of f
we know that ∃δ>0:a∈Bδ(x)⟹f(a)∈Bϵ(f(x))⊂V
Thus Bδ(x)⊂f−1(V)
(as for all a
in the ball, the thing f
maps it to is in the ball of radius ϵ
about f(x)
).
Since x
was arbitrary we have ∀x∈f−1(V)∃an open ball containing x⊂f−1(V)
, thus f−1(V)
is open.
⟸
Choose any x∈X
Let ϵ>0
be given.
As Bϵ(f(x))
is an open set, the hypothesis implies that f−1(Bϵ(f(x)))
is open in X
Since x∈f−1(Bϵ(f(x)))
and f−1(Bϵ(f(x)))
is open, it is a neighborhood to all of its points, that means
∃δ>0:Bδ(x)⊂f−1(Bϵ(f(x)))
Note: we have now shown that ∀ϵ>0∃δ>0:Bδ(x)⊂f−1(Bϵ(f(x)))
Using the implies and subset relation we see a∈Bδ(x)⟹a∈f−1(Bϵ(f(x))) which then ⟹f(a)∈Bϵ(f(x))
Or just a∈Bδ(x)⟹f(a)∈Bϵ(f(x)))
Thus it is continuous at x
, since x
was arbitrary, it is continuous.
