Suppose that #T# is continuous and that #AsubeY# is any open set in #Y#. We need to show that #T^(−1)(A)# is open.
Let #x in T^(−1)(A)# be arbitrary.
Then, #y = T(x) in A# and hence there exists #epsilon > 0# such that #B(y,epsilon) in A# (since #A# is open).
Since #T# is continuous, there exists #delta > 0# such that #T(B(x,delta)) sube B(y,epsilon) sube A#. So #B(x,delta) sube T^(−1)(A)#.
We have shown that for every point #x in T^(−1)(A)# there is an open ball centered at #x# that is contained in #T^(−1)(A)#. So, #T^(−1)(A)# is an open set.
Assume that the inverse image of any open set is open.
Let #x in X, epsilon > 0# be arbitrary, and let #B(y,epsilon)# be an open ball centered at #y = T(x) in Y# of radius #epsilon#.
The inverse image #T^(−1)(B(y,epsilon))# of the open ball is an open set that contains the point #x# (since #y = T(x)# is certainly a member of #B(y,epsilon)#).
Therefore there exists a ball #B(x,delta)# about #x# such that #B(x,delta) sube T^(−1)(B(y,epsilon))# or, equivalently, #T(B(x,delta)) sube B(y,epsilon)#.
Since #x, epsilon# were arbitrary, it follows that #T# is continuous.
