Is Sequence Convergence a topological Invariant?

Question

I am stuck on a question that goes like this:

Let $(X, \mathcal{T})$ be a topological space, and let $A \subseteq X$. $A$ is called sequentially closed if the limit point of every convergent sequence $\{x_n\} \subseteq A$ is in $A$. We already know that every closed set is sequentially closed. $(X, \mathcal{T})$ is called sequential if every sequentially closed subset is closed. Show that the property of being sequential is a topological invariant.

I have a solution but that relies on the postulate that sequence convergence is a topological invariant. Here is my proof:

Let's suppose we have two topological spaces $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ that are homeomorphic, with a homeomorphism $f: X \to Y$. We want to show that if $(X, \mathcal{T})$ is sequential, then $(Y, \mathcal{U})$ is also sequential, and vice versa.

First, let's prove that if $(X, \mathcal{T})$ is sequential, then $(Y, \mathcal{U})$ is sequential. We need to show that every sequentially closed subset $B \subseteq Y$ is closed in $(Y, \mathcal{U})$

Suppose $B$ is a sequentially closed subset of $(Y, \mathcal{U})$. We want to prove that $B$ is closed in $(Y, \mathcal{U})$. Consider the preimage $A = f^{-1}(B) \subseteq X$. Since $f$ is a homeomorphism, $A$ is a subset of $X$, and we have $f(A) = B$

Now, let ${y_n}$ be a convergent sequence in $B$ with limit point $y$. Since $f$ is a homeomorphism, the preimage ${x_n} = f^{-1}({y_n})$ is a convergent sequence in $A$ with limit point $x = f^{-1}(y)$. Since $A$ is sequentially closed in $(X, \mathcal{T})$, $x$ belongs to $A$. By continuity of $f$, we have $f(x) = y$. Thus, $y$ is in the image of $A$ under $f$, which is $B$. Therefore, $B$ contains its limit point $y$, and $B$ is closed in $(Y, \mathcal{U})$. Hence, if $(X, \mathcal{T})$ is sequential, then $(Y, \mathcal{U})$ is also sequential

The converse that $(X, \mathcal{T})$ is sequential if $(Y, \mathcal{U})$ is can also be proven in the same way as there was nothing special about $(Y, \mathcal{U})$. Therefore, we have shown that the property of being sequential is a topological invariant.

Any comments on my proof/idea is welcome. I need help in solving this problem.

Answer 1

Your idea to set up a homeomorphism is good. But you are confusing a few concepts.

Let's start at this point in your proof (slightly paraphrased):

Suppose $B \subseteq Y$ is sequentially closed. We want to prove that $B$ is closed.

You proceeded to select a sequence $\{ y_n \}_{n \in \mathbb{Z}^+} \subseteq B$ that converges to $y$. Then, you showed that $y \in B$. But this is given by hypothesis that $B$ is sequentially closed and doesn't actually show that $B$ is closed.

Let's consider the set $A = f^{-1} (B)$. We show that $A$ is sequentially closed. To see this take any sequence $\{ a_n \}$ of elements of $A$ that converges to $a \in X$. Take the sequence $\{ f (a_n) \}$ of elements of $B$. Let $V$ be an arbitrary neighborhood of $f (a)$ in $Y$. Since $f$ is continuous, $f^{-1} (V)$ is open in $X$; indeed, it is a neighborhood of $a$ in $X$, so there exists $N \in \mathbb{Z}^+$ such that $a_n \in f^{-1} (V)$ for all $n > N$. This means of course that $f (a_n) \in V$ for all $n > N$. Since $V$ was arbitrary, $\{ f (a_n) \}$ converges to $f (a)$. We had by hypothesis that each $a_n$ is contained in $A$. Therefore, each $f (a_n)$ is contained in $B$. The sequence $\{ f (a_n) \}$ of elements of $B$ converges to $f (a)$, and since $B$ is sequentially closed, $f (a) \in B$, so that $a \in A$, as desired.

By hypothesis, $X$ is sequential, so the fact that $A$ is sequentially closed means that it is also closed.

Finally, we note that since $f$ is a homeomorphism, it is a closed map. So, $B$, being the image of a closed set, is also closed.

Answer 2

Suppose $\{x_n\}$ is a convergent sequence in $A$, $x$ is the limit of $\{x_n\}$, now we want to prove $x\in A$. It suffices to prove that $f(x)\in B$. And it suffices to prove that $f(x)$ is a limit of $f(x_n)$.

Let's take arbitrary open neighborhood $O$ of $f(x)$, then $x\in f^{-1}(O)$ and $f^{-1}(O)$ is open in $X$. Since $x_n\to x$, there exists $N$ such as $\forall n\geq N, x_n\in f^{-1}(O) $, which means $f(x_n)\in O,\forall n\geq N$. Hence $f(x)$ is a limit of $f(x_n)$ by definition.