Thread: real hard problem

Originally Posted by mancillaj3

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.

Originally Posted by mancillaj3

Consider a function . Prove is continuous on iff for every sunbset A of

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