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I have questions about the relation between continuity and interior based on the article ;A map is continuous if and only if for every set, the image of closure is contained in the closure of image

At first I guess that there will be a property like $f:X\rightarrow Y$ is continuous if and only if $\forall A\subset X,\ f(A)^{i} \subset f(A^{i}) $ where $A^{i}$ denotes the interior of $A$.

However, I found some counter examples ;


Eg.1 for $(\Rightarrow)$

Define $f:\mathbb{R} \rightarrow \mathbb{R}$

by $ f(x) = \left\{ \begin{array}{l l} x & \quad \text{if $x \in(-\infty,1]$}\\ 1 & \quad \text{if $x \in[1,2]$}\\ x-1 & \quad \text{if $x \in[2,\infty)$} \end{array} \right.$

then $f$ is continuous

put $A=[0,1]\cup [2,3]\subset \mathbb{R}$

Then, we can observe

$f(A)=[0,2]\Rightarrow f(A)^{i}=(0,2)$

$A^{i}=(0,1)\cup (2,3)\Rightarrow f(A^{i})=[0,1)\cup(1,2]$

Hence, $f(A)^{i} \nsubseteq f(A^{i}) $


Eg.2 for $(\Leftarrow)$ Define $f:\mathbb{R} \rightarrow \mathbb{R}$ by $f(x)=\bigg\{^{x+1 \ \ if\ \ x\in [0,\infty)}_{x \ \ if\ \ x\in (-\infty,0)}$

then $f$ is not continuous at $x=0$, but the interior part statement holds for all subset of its domain.

Suppose $\exists A\subset \mathbb{R} $ s.t. $f(A)^{i}-f(A^{i}) \not=\emptyset $

put $ y\in f(A)^{i}-f(A^{i})$

if $y=1, \exists \ \epsilon >0 $ s.t. $(1-\epsilon,1+\epsilon)\subset f(A)$

Which means, $\ \max\{\frac{1}{2},1-\frac{\epsilon}{2}\}\in (1-\epsilon,1+\epsilon)\subset f(A) \subset \mathbb{R}-[0,1)$ ;contradiction

if $y<1$, then $ y<0 $

$\implies \exists \ \epsilon >0 $ s.t. $(y-\epsilon,y+\epsilon)\subset f(A) $

$\implies (y-\epsilon,y+\epsilon)\subset A $ $\implies y\in A^{i}$ $\implies y\in f(A^{i})$ ; contradiction

if $ y>1, \exists \ \epsilon >0$ s.t. $(y-\epsilon,y+\epsilon)\subset f(A)$

then we can assume $y-\epsilon >1$ which implies $(y-\epsilon -1,y+\epsilon -1)\subset A$

$\implies y-1 \in A^{i} $ $\implies y\in f(A^{i})$ ; contradiction

Hence this can be a counter example for $(\Leftarrow)$


However, in these examples I assumed that the interior of a subset in the codomain is defined with respect to a topology of $\mathbb{R}$, not $f(\mathbb{R})$. This is because when we check the continuity between topology spaces, we just consider whether the inverse of a open set in the codomain is open or not. but I'm not sure if such an assumption makes sense...

So I changed the statement to $f(A)^{i} \supset f(A^{i})$. For this, one can easily find a counter example for $(\Rightarrow)$ if $f(x)=\sin (x)$, but I have not been able to make progress on the other direction. Could anyone give me a hint for this?

Sorry for reading these long questions

summary;

  1. if the assumption I used in the example makes loss of generality or not.

  2. proof(any hint) or counter example for $f:X\rightarrow Y$ is continuous if $\forall A\subset X, f(A)^{i} \supset f(A^{i})$.

  3. (Extra) If there is any property between the continuity and the interior, please let me know (not for the inverse image).

Thank you again for reading all.

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    $\begingroup$ The property $f(A^i)\subseteq f(A)^i$ means that the map is open, and there are open maps which are not continuous and vice versa. If you take the interior relative to $f(X)$, then this means that the function $f:X\to f(X)$ is open. $\endgroup$ Mar 27 '15 at 17:55
  • $\begingroup$ I think the map $\sin(x)$ is open as a map $\Bbb R\to[-1,1]$. $\endgroup$ Mar 27 '15 at 18:01
  • $\begingroup$ I didn't check that point. So if a continuous bijection satisfies that, then we can conclude it is a homeomorphism. Thank you for your comment :) $\endgroup$
    – alemonk
    Mar 28 '15 at 5:12
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You're absolutely right with counter-examples that you gave above, however there is a corresponding theorem, not on $f$, but on the preimage of sets:

If $X$ and $Y$ are topological spaces then $f$ is continuous if and only if $f^{-1}(A^i) \subseteq (f^{-1}(A))^i$ for all $A \subseteq Y$.

Proof: ($\Longrightarrow$) Suppose $f$ is continuous and let $p \in f^{-1}(A^i)$. This implies that $f(p) \in A^i$ hence there is an open neighborhood $U \subseteq A^i$ containing $f(p)$. Certainly $U \subseteq A$ therefore $p \in f^{-1}(U) \subseteq f^{-1}(A)$. Since $f$ is continuous we have that $f^{-1}(U)$ is an open neighborhood of $p$ in $f^{-1}(A)$, hence $p \in (f^{-1}(A))^i$.

($\Longleftarrow$) Now suppose $f:X\to Y$ satisfies the above property, and let $U \subseteq Y$ be open. We have that $f^{-1}(U^i) = f^{-1}(U) \subseteq (f^{-1}(U))^i \subseteq f^{-1}(U)$. Therefore $f^{-1}(U) = (f^{-1}(U))^i$, and so the preimage of open subsets of $Y$ under $f$ are open in $X$. This proves that $f$ is continuous.

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