Proof. We do the easy direction first. Suppose f is continuous at x 0 and let t be a fixed positive number. There is a 8 > 0 such that if llx- xoll < 8 then llf(x)- f(xo)ll < t. Furthermore, if {xn} is a sequence with limit x 0 then there is anN > 0 such that llxn- xoll < 8 Some simple results 19 for all n 2:: N. ( xn)} to have limit J( x 0 ). The other half of the proof is effected by a 'contrapositive' argument.

Again, simply by rewriting this symbolically, we see that llf(x)- f(xo)ll < f whenever llx- xoll < 8 (which is what it means to say xis in Bs(xo)). Thus f is indeed continuous. We could now show that (i) and (iv) are equivalent by arguments modelled on but different from those above. This would be an instructive exercise, but by now the idea of using set complementation should spring easily to mind. The starting point is 1 (Y \G) =X\ f- 1 (G). The to observe that for any set Gin Y, details are left to the reader.

10 immediately. If Xn --+ x 0 implies f(xn) --+ f(x 0 ) then we say f is sequentially continuous at x. However, the concepts of continuity and sequential continuity are not always equivalent. 8 but this definition is not equivalent to sequential continuity. 4: The function f(x) = 1/x near x = 0. very small as we shift our attention from point to point in X. This underlines the fact that the parameter 6 is dependent on the point x in the definition. For example, consider the case X = Y = 1R with the norm llxll = lxl.

### An Introduction to Abstract Analysis by W. A. Light (auth.)

by Michael

4.1