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 $\displaystyle \mathbb{R}$ $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$
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).
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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, $\displaystyle f(\overline{A})\subseteq\overline{f(A)}$.
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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.
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
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