Let $(X, d_X)$, $(Y, d_Y)$ be metric spaces and $f: X \to Y$ be a function. The metric $d_{X \times Y}((x_1, y_1), (x_2, y_2)) = d_X(x_1, x_2) + d_Y(y_1, y_2)$ defines the Cartesian product $X \times Y$. The graph of $f$ is defined as $G(f) = {(x, f(x)) : x \in X}$.
Suppose $(X, d)$ is sequentially compact. Show that $f$ is continuous if and only if $G(f) \subseteq X \times Y$ is sequentially compact.
$\textbf{Mysolution:}$
$\textit{Part 1: Assuming $f$ is continuous, prove that $G(f)$ is sequentially compact.}$
Let $(x_n, f(x_n))$ be a sequence in $G(f)$. Since $G(f) = \{(x, f(x)) : x \in X\}$, each element $(x_n, f(x_n))$ corresponds to some $x_n \in X$.
Since $X$ is sequentially compact, there exists a convergent subsequence $(x_{n_k})$ that converges to some $x \in X$.
Now, let's consider the corresponding subsequence $(f(x_{n_k}))$ in $Y$. Since $f$ is continuous, we know that $f(x_{n_k})$ converges to $f(x)$ as $k$ approaches infinity.
Therefore, we have shown that for any sequence $(x_n, f(x_n))$ in $G(f)$, there exists a convergent subsequence that converges to some point $(x, f(x))$ in $G(f)$.
Hence, $G(f)$ is sequentially compact.
\textit{Part 2: Assuming $G(f)$ is sequentially compact, prove that $f$ is continuous.}
Let $x_n$ be a sequence in $X$ that converges to some $x \in X$. We need to show that $f(x_n)$ converges to $f(x)$ in $Y$.
Consider the sequence $(x_n, f(x_n))$ in $G(f)$. Since $G(f)$ is sequentially compact, there exists a convergent subsequence $(x_{n_k}, f(x_{n_k}))$ that converges to some point $(x', y')$ in $G(f)$.
Since $(x_n, f(x_n))$ converges to $(x', y')$ in $G(f)$, it follows that $x_n$ converges to $x'$ in $X$ and $f(x_n)$ converges to $y'$ in $Y$.
But $x_n$ also converges to $x$ in $X$. By the uniqueness of limits, we have $x = x'$.
Since $f(x_n)$ converges to both $y'$ and $f(x)$, we conclude that $y' = f(x)$.
Therefore, $f(x_n)$ converges to $f(x)$ in $Y$.
Hence, we have shown that if $G(f)$ is sequentially compact, then $f$ is continuous.
In summary, we have proven that $f$ is continuous if and only if $G(f)$ is sequentially compact when $X$ is sequentially compact.
what about my solution, is that correct? I'm not that satisfied with my proof