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Suppose $f:\mathbb{R}\longrightarrow \mathbb{R}$ be an arbitrary continuous fuction such that the inverse image of a bounded set is bounded. Then show that,

$1$) The image under $f$ of a closed set is closed.

$2$) $f$ is not necessarily a surjective function.

My attempt ($1$) :

Say, $X\subset \mathbb{R}$ is an arbitrary closed set, such that $f(X)=Y\subset \mathbb{R}$ is not closed. Then I am trying to prove by contradiction.
Case 1 : $X$ is bounded. Hence $X$ is closed and bounded $\implies$ compact. Since $f$ is continuous, $f(X)=Y$ is also compact $\implies Y$ is closed. (a contradiction)
Case 2 : $X$ is not closed and not bounded. So $f(X)=Y$ is not bounded and not closed. If $Y$ is open and not bounded then $Y=\mathbb{R}$, but $\mathbb{R}$ is clopen (closed and open). (a contradiction).
Hence, $f(X)=Y$ where $X$ is closed and not bounded and $Y$ is neither open nor closed and also not bounded. Now I am confused that how to get a contradiction of this final case.

My attempt ($2$) :

Here, I cannot understand the meaning of the inverse image of a bounded set. I think inverse function will exist only for the bijections. In this context how I can find a continuous map $f$ which is not a surjection, and inverse image of the bounded set is bounded.

Shankhadeep
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    For (2), "the inverse image of a bounded set is bounded" most likely means "the pre-image of a bounded set in the co-domain, which is a subset of the domain, is also bounded". – Mahmoud Dec 20 '21 at 08:38
  • Ok ! Then I think a constant function can act as an example, right ? – Shankhadeep Dec 20 '21 at 08:41
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    $(2)$ $\Bbb R\ni x\longmapsto x^2\in \Bbb R$ – Sumanta Dec 20 '21 at 08:43
  • It seems to me you have difficulties since you are confusing inverse image (of a generic function) and inverse function (which exists if and only if the map is a bijection). Try to prove first that not always it is the case that $f^{-1}f(X)= X$ and try to understand why. Then some of these passages will be clearer. – Son Gohan Dec 20 '21 at 08:43
  • Yeah , I got the point but still I am confused for the first part i.e closed implies closed. – Shankhadeep Dec 20 '21 at 08:45
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    A constant function may not be the best example, since the image of every subset of the domain consists of just one element. This would mean that an unbounded set in the domain would also have a bounded image. The function $f(x) = x^2$ would probably better. – Mahmoud Dec 20 '21 at 08:45
  • ok , thanks, but can you please help me in the first part ? – Shankhadeep Dec 20 '21 at 08:47
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    $(1)$ Let $X\subseteq_\text{closed}\Bbb R$ and ${x_n}\subseteq X$ with $f(x_n)\to y\in \Bbb R$. Now, all but finitely many $x_n\in f^{-1}\big((y-1,y+1)\big)=$ a bounded set. So, ${x_n}$ is a bounded sequence. As $X$ is closed each limit point of ${x_n}$ is in $X$ (recall Bolzano-Weierstrass Theorem: Every bounded sequence has limit point). Let $\ell$ be a limit point of ${x_n}$ and $x_{n_k}\to \ell$, then $f(x_{n_k})\to f(\ell)$ as $f$ is continuous. Also, $f(x_{n_k})\to y$, i.e., $y=f(\ell)\in f(X).$ – Sumanta Dec 20 '21 at 08:52

2 Answers2

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If $C\subset \mathbb R$ is closed and bounded then $f(C)$ is closed and bounded too.

So let $C\subset \mathbb R $ be closed and unbounded. Suppose that $f(C)$ is the image of $C$ under $f$.

Let $p$ be a given limit point of $f(C)$. For any $n\in \mathbb N$, there is an $x_n\in C$ such that $f(x_n)\in (p-1/n, p+1/n)$. So $f(x_n)\to p$.

It follows that the sequence $(f(x_n))$ is bounded and hence $(x_n)$ is also bounded (by given hypothesis). By Bolzanno Weierstrass theorem, there exists a convergent subsequence $x_{n_k}\to l\in C$ (because $C$ is closed) hence by continuity, $f(x_{n_k})\to f(l)=p$. It follows that $p\in f(C)$, hence $f(C)$ is closed.

For $(2)$, consider $f(x)=e^{|x|}$. $f(x)\ge 1$ for all $x\in \mathbb R$.

Koro
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If $f: X\to Y$ is an arbitrary map, the inverse image of a subset $Y'$ of $Y$ is defined as $f^{-1}(Y'):=\{x\in X: f(x)\in Y'\}$. This dose not require $f$ to be bijective.

For (1), let $X$ be a closed subset of $\mathbb{R}$, then for any $y$ lies in the boundary of $f(X)$, by the definition, you can take a sequence of points $y_n$ in $f(X)$, such that $\lim_{n\to\infty}y_n=y$, and for each $y_n$, take $x_n\in X$ such that $f(x_n)=y_n$. $\{y_n: n\}$ is bouned, hence $\{x_n: n\}$ is also bounded, then pick a convergent subsequence of $(x_n)_n$ (still denoted by $(x_n)_n$), and set $x=\lim_{n\to\infty}x_n$. $X$ is closed, hence $x\in X$. $f$ is continuous, hence $f(x)=f(\lim_{n\to\infty}x_n)=\lim_{n\to\infty}f(x_n)=y$, that is, $y\in f(X)$. Therefore, $f(X)$ is also closed. As for (2), $f(x)=\mathrm{exp}(x)$ should be an example.

DU CJ
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