Consider a function . Prove is continuous on iff for every sunbset A of
Lemma 1. f is continuous iff for each closed subset C of (codomain of f), is closed in (domain of f).
->
Assume f is continuouus and let A be a subset of (domain of f). Then, is a closed subset of (codomain of f), so its inverse is closed in (domain of f) by lemma 1.
Since (the latter set is closed), we have .
Thus, .
<-
Assume is true for each subset A of (domain of f).
Let C be a closed subset of (codomain of f). Then, . Thus, , which implies that is a closed subset in (domain of f).
We conclude that f is continuous by lemma 1.
Here is another proof:
For that among other theorems we will use the following theorem:
Given any subset A of the real Nos R AND xεR,THEN:
iff there exists a sequence in A and such that
Now let =====> THERE exists a sequence in A and
BUT f is also continuous over R AND hence over A AND thus
Also is in A HENCE is in f(A) AND so
Thus
Conversely:
Suppose and also that f is not continuous over R.
Then there exist xεR AND a sequence in R such that and .................................................. ............................1.
Now put ,hence .
BUT by the construction of A WE have:
.................................................. ...........................2
(1) and (2) lead us to a contradiction,hence f is continuous over R
I tried to give another proof by using the ε-δ definitions only,so
Let ,hence x belongs to the real Nos R
Let ε>0
Since f is continuous over R AND thus at xεR we have :
There exists δ>0 and such that:
, .................................................. ........................1
Since ,for any +ve No and thus for δ>0,there exists a ,y and such that:
yεB(x,δ) and yεA,or
|x-y|<δ and yεΑ............................................... ............................2
put now z=y in (1) and
and since yεA =====> f(y)εf(A) WE have ,by using (2):
|f(x)-f(y)|<ε and f(y)εf(A) ,SO
FOR all ε>0 ,then f(y)εB(f(x),ε) and f(y)εf(A).HENCE
.
For the converse i could not find a proof without using sequences or closed sets