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


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?


  • 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

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.


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