1. ## real hard problem

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

2. Originally Posted by mancillaj3
Consider a function $f:\mathbb{R}\rightarrow\mathbb{R}$. Prove $f$ is continuous on $\mathbb{R}$ iff for every sunbset A of $\mathbb{R}$ $f(\overline{A})\subseteq\overline{f(A)}$
Lemma 1. f is continuous iff for each closed subset C of $\mathbb{R}$ (codomain of f), $f^{-1}(C)$ is closed in $\mathbb{R}$ (domain of f).

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

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Assume $f(\overline{A})\subseteq\overline{f(A)}$ is true for each subset A of $\mathbb{R}$ (domain of f).
Let C be a closed subset of $\mathbb{R}$ (codomain of f). Then, $f(\overline{f^{-1}(C)}) \subset \overline{ff^{-1}(C)} \subset \overline{C}=C$. Thus, $\overline{f^{-1}(C)} \subset f^{-1}(C)$, which implies that $f^{-1}(C)$ is a closed subset in $\mathbb{R}$ (domain of f).
We conclude that f is continuous by lemma 1.

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

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

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

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

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

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

Conversely:

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

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

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

BUT by the construction of A WE have:

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

(1) and (2) lead us to a contradiction,hence f is continuous over R

4. I tried to give another proof by using the ε-δ definitions only,so

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

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

Since $x\in\overline A$ ,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

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

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

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