3
$\begingroup$

Definition 1: $A \subset (X, \mathcal{T})$ is sequentially closed if the limit of all convergent sequence $(x_n)$ in $A$ is in $A$.

Definition 2: $(X, \mathcal{T})$ is a sequential topological space if every sequentially closed subset is closed.

Definition 3: A set is closed if its complement is open


Show that if $(X, \mathcal{T})$ is a sequential topological space , and $f: (X, \mathcal{T}) \to (Y, \mathcal{J})$ is a homeomorphism then $(Y, \mathcal{J})$ is a sequential topological space


Attempt:

Let $(X, \mathcal{T})$ be a sequential topological space, and let $f: (X, \mathcal{T}) \to (Y, \mathcal{J})$ be a homeomorphism. Let $A \subset X$ be a sequentially closed set, then the limit of all convergent sequences is in $A$. Furthermore $A$ is closed by definition of $X$.

We know that homeomorphism (in particular, continuous map) preserves convergence of sequence, then given a sequence $(x_n)$ in $A$, the image sequence $(f(x_n))$ is a convergent sequence in $B := f(A)$. Since a homeomorphism is a continuous closed map, therefore $B$ is closed.

Since $f$ maps convergent sequence to convergent sequence, and preserves closedness, therefore $B \subset (Y, \mathcal{J})$ is a sequentially closed topological space.


Attempt 2:

Let $B \subset Y$ be a sequentially closed set. Then given convergent sequence $(y_n)$ in $B, y_n \to y \in B$.

Then we wish to prove that $B$ is closed.

Let $X$ be a sequential topological space and $f$ be a homeomorphism between $X$ and $Y$, then $f^{-1}$ is continuous. Thus $(f^{-1}(y_n))$ is a convergent sequence in $f^{-1}(B)$. This shows $f^{-1}(B)$ is sequentially closed. By definition of $X$, $f^{-1}(B)$ is closed.

Since $f$ sends closed sets to closed sets, $f(f^{-1}(B)) = B$ is closed, therefore sequential topological spaces are preserved under homeomorphism.

**

Can someone check whether attempt 2 is correct?

**

$\endgroup$
  • 2
    $\begingroup$ You've proved that $f$ maps a sequentially closed set $A$ to a closed set $B$. You have not proved that every sequentially closed subset of $Y$ is closed in $Y$. You should start by picking a sequentially closed subset of $Y$, and show that $f^{-1}$ maps it back to a sequentially closed set. And be aware that in non-first-countable spaces, a sequentially closed set is not necessarily closed. $\endgroup$ – Qiyu Wen Jun 14 '16 at 23:11
  • $\begingroup$ @QiyuWen Thanks, I thought the proof was awkward...let me start over $\endgroup$ – Carlos - the Mongoose - Danger Jun 14 '16 at 23:13
1
$\begingroup$

I think that the idea in attempt 2 is good. But if you want to prove that $f^{-1}(B)$ is sequentially closed, you have to say that the limit of every convergent sequence is in $f^{-1}(B)$. Thus, take a convergent sequence $\{x_n\}_{n\in\mathbb{N}}$in $f^{-1}(B)$ which converges to $x$. Then, since $f$ is continuous, $f(x_n)$ converges to $f(x)$. Since $\{f(x_n)\}_{n\in\mathbb{N}}\subseteq B$ and $B$ sequentially closed $f(x)\in B$. Therefore, $x\in f^{-1}(B)$. Hence, it is sequentially closed. Then, you can conclude as you did.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.