# [SOLVED]Continuous Map Property

#### Sudharaka

##### Well-known member
MHB Math Helper
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
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.

Am I doing something stupid?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

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
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
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
Thanks caffeinemachine and I like Serana. I truly appreciate your help.

So then what if I relax my condition to,

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

#### Opalg

##### MHB Oldtimer
Staff member
Thanks caffeinemachine and I like Serana. I truly appreciate your help.

So then what if I relax my condition to,

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

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 $[f(A)]' = \bigl[-\frac\pi2,\frac\pi2\bigr] \not\subseteq f(A').$
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 $[f(A)]' = \bigl[-\frac\pi2,\frac\pi2\bigr] \not\subseteq f(A').$