# real hard problem

• Apr 18th 2009, 11:30 PM
mancillaj3
real hard problem
Consider a function $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$. Prove $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$ iff for every sunbset A of $\displaystyle \mathbb{R}$ $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$
• Apr 19th 2009, 07:32 PM
aliceinwonderland
Quote:

Originally Posted by mancillaj3
Consider a function $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$. Prove $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$ iff for every sunbset A of $\displaystyle \mathbb{R}$ $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$

Lemma 1. f is continuous iff for each closed subset C of $\displaystyle \mathbb{R}$ (codomain of f), $\displaystyle f^{-1}(C)$ is closed in $\displaystyle \mathbb{R}$ (domain of f).

->
Assume f is continuouus and let A be a subset of $\displaystyle \mathbb{R}$ (domain of f). Then, $\displaystyle \overline{f(A)}$ is a closed subset of $\displaystyle \mathbb{R}$ (codomain of f), so its inverse $\displaystyle f^{-1}(\overline{f(A)})$ is closed in $\displaystyle \mathbb{R}$ (domain of f) by lemma 1.
Since $\displaystyle A \subset f^{-1}(\overline{f(A)})$ (the latter set is closed), we have $\displaystyle \overline{A} \subset f^{-1}(\overline{f(A)})$.
Thus, $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$.

<-
Assume $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$ is true for each subset A of $\displaystyle \mathbb{R}$ (domain of f).
Let C be a closed subset of $\displaystyle \mathbb{R}$ (codomain of f). Then, $\displaystyle f(\overline{f^{-1}(C)}) \subset \overline{ff^{-1}(C)} \subset \overline{C}=C$. Thus, $\displaystyle \overline{f^{-1}(C)} \subset f^{-1}(C)$, which implies that $\displaystyle f^{-1}(C)$ is a closed subset in $\displaystyle \mathbb{R}$ (domain of f).
We conclude that f is continuous by lemma 1.
• Apr 22nd 2009, 05:54 PM
xalk
Quote:

Originally Posted by mancillaj3
Consider a function $\displaystyle f:\mathbb{R}\rightarrow\mathbb{R}$. Prove $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$ iff for every sunbset A of $\displaystyle \mathbb{R}$ $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$

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:

$\displaystyle x\in\overline A$ iff there exists a sequence $\displaystyle \{x_{n}\}$ in A and such that $\displaystyle \lim_{n\rightarrow\infty}{x_{n}}= x$

Now let $\displaystyle y\in f(\overline A)\Longrightarrow y=f(x)\wedge x\in\overline A$=====> THERE exists a sequence $\displaystyle \{x_{n}\}$ in A and $\displaystyle \lim_{n\rightarrow\infty}{x_{n}}=x$

BUT f is also continuous over R AND hence over A AND thus $\displaystyle \lim_{n\rightarrow\infty}{f(x_{n})} = f(x)$

Also $\displaystyle \{x_{n}\}$ is in A HENCE $\displaystyle f(x_{n})$ is in f(A) AND so $\displaystyle f(x)\in\overline {f(A)}$

Thus $\displaystyle f(\overline A)\subset\overline {f(A)}$

Conversely:

Suppose $\displaystyle f(\overline A)\subset\overline {f(A)}$ and also that f is not continuous over R.

Then there exist xεR AND a sequence in R $\displaystyle \{x_{n}\}$ such that $\displaystyle \lim_{n\rightarrow\infty}{x_{n}}=x$ and $\displaystyle \lim_{n\rightarrow\infty}{f(x_{n})}\neq f(x)$.................................................. ............................1.

Now put $\displaystyle A= \{x_{n}: n\in N\}=\{x_{n}\}$,hence $\displaystyle x\in\overline A\Longrightarrow f(x)\in\ f(\overline A)\Longrightarrow f(x)\in\overline{f(A)}$.

BUT by the construction of A WE have:

$\displaystyle \lim_{n\rightarrow\infty}{f(x_{n})}=f(x)$.................................................. ...........................2

(1) and (2) lead us to a contradiction,hence f is continuous over R
• Apr 25th 2009, 09:54 AM
xalk
I tried to give another proof by using the ε-δ definitions only,so

Let $\displaystyle w\in f(\overline A)\Longrightarrow w=f(x) ,x\in\overline A$,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:

$\displaystyle \forall z$, $\displaystyle |z-x|<\delta\Longrightarrow |f(x)-f(z)|<\epsilon$.................................................. ........................1

Since $\displaystyle x\in\overline A$ ,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

$\displaystyle |x-y|<\delta\Longrightarrow |f(x)-f(y)|<\epsilon$ 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

$\displaystyle w=f(x)\in\overline{f(A)}$.

For the converse i could not find a proof without using sequences or closed sets