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[SOLVED] Continuous Map Property

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi everyone, :)

Trying hard to do a problem recently, I encountered the following question. Hope you can shed some light on it. :)

Suppose we have a continuous mapping between two metric spaces; \(f:\, X\rightarrow Y\). Let \(A\) be a subspace of \(X\). Is it true that,

\[f(A')=[f(A)]'\]

where \(A'\) is the set of limit points of \(A\) and \([f(A)]'\) is the set of limit points of \(f(A)\).
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi everyone, :)

Trying hard to do a problem recently, I encountered the following question. Hope you can shed some light on it. :)

Suppose we have a continuous mapping between two metric spaces; \(f:\, X\rightarrow Y\). Let \(A\) be a subspace of \(X\). Is it true that,

\[f(A')=[f(A)]'\]

where \(A'\) is the set of limit points of \(A\) and \([f(A)]'\) is the set of limit points of \(f(A)\).
Let $X=\mathbb R$ and $Y=\{y\}$ be a singleton. Take $A=(0,1)$. Define $f:X\to Y$ as $f(x)=y$ for all $x\in X$. Then $f$ is continuous. Now $A$ has limit points but $f(A)$ has none. So $f(A')=[f(A)]'$ cannot hold.

(Wait) Am I doing something stupid?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Let $X=\mathbb R$ and $Y=\{y\}$ be a singleton. Take $A=(0,1)$. Define $f:X\to Y$ as $f(x)=y$ for all $x\in X$. Then $f$ is continuous. Now $A$ has limit points but $f(A)$ has none. So $f(A')=[f(A)]'$ cannot hold.

(Wait) Am I doing something stupid?
I think so.
$f(A')=f([0,1])=\{y\}$
$[f(A)]'=f((0,1))'=\{y\}'=\{y\}$
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
I think so.
$f(A')=f([0,1])=\{y\}$
$[f(A)]'=f((0,1))'=\{y\}'=\{y\}$
This means $y$ is the limit point of $\{y\}$.
I think this is incorrect. For the above to be true, any open set in $Y$ which contains $y$ should contain a point of $\{y\}$ which is different from $y$. This clearly isn't true.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
This means $y$ is the limit point of $\{y\}$.
I think this is incorrect. For the above to be true, any open set in $Y$ which contains $y$ should contain a point of $\{y\}$ which is different from $y$. This clearly isn't true.
Good point.
I mixed up closure (often denoted with a prime) with the set of limit points.
So indeed:

$f(A')=f([0,1])=\{y\}$
$[f(A)]'=f((0,1))'=\{y\}'=\varnothing$

Therefore the statement in the OP is not true.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Thanks caffeinemachine and I like Serana. I truly appreciate your help. :)

So then what if I relax my condition to,

\[[f(A)]'\subseteq f(A')\]

What do you think about this? Any counter examples? :)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
Thanks caffeinemachine and I like Serana. I truly appreciate your help. :)

So then what if I relax my condition to,

\[[f(A)]'\subseteq f(A')\]

What do you think about this? Any counter examples? :)
Take $A = X = Y = \mathbb{R}$ and let $f(x) = \arctan x$. Then $A' = A$, and $f(A') = f(A) = \bigl(-\frac\pi2,\frac\pi2\bigr)$. But $[f(A)]' = \bigl[-\frac\pi2,\frac\pi2\bigr] \not\subseteq f(A').$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Take $A = X = Y = \mathbb{R}$ and let $f(x) = \arctan x$. Then $A' = A$, and $f(A') = f(A) = \bigl(-\frac\pi2,\frac\pi2\bigr)$. But $[f(A)]' = \bigl[-\frac\pi2,\frac\pi2\bigr] \not\subseteq f(A').$
Thank you so much. That indeed is a nice and simple example, just what I have been looking for. :)