[proofplan] The only point is that the subspace metric does not change distances between points of $Y$. We write out the $\varepsilon$ definition of convergence in $(Y,d_Y)$, replace $d_Y(y_k,y)$ by $d(y_k,y)$, and observe that the resulting condition is exactly the $\varepsilon$ definition of convergence in $(X,d)$. The reverse implication uses the same equality of distances. [/proofplan]

[step:Identify the two convergence conditions as the same eventual inequality]By the definition of the subspace metric, equivalently by [citetheorem:8585], the map \begin{align*} d_Y:Y\times Y\to [0,\infty) \end{align*} satisfies \begin{align*} d_Y(a,b)=d(a,b) \end{align*} for every $a,b\in Y$. Since $(y_k)_{k=1}^{\infty}$ is a sequence in $Y$ and $y\in Y$, we have $y_k\in Y$ and $y\in Y$ for every $k\in \mathbb N$. Hence, for every $k\in \mathbb N$, \begin{align*} d_Y(y_k,y)=d(y_k,y). \end{align*}[/step]

[guided]The subspace metric is built by restricting the ambient metric to pairs of points that lie in $Y$. More explicitly, the map \begin{align*} d_Y:Y\times Y\to [0,\infty) \end{align*} is given by \begin{align*} d_Y(a,b)=d(a,b) \end{align*} for all $a,b\in Y$. This equality is the whole mechanism of the proof. The sequence hypothesis says that each term $y_k$ belongs to $Y$, and the candidate limit also satisfies $y\in Y$. Therefore the pair $(y_k,y)$ lies in $Y\times Y$ for every $k\in \mathbb N$, so the defining equality for the restricted metric applies to this pair. Thus, for every $k\in \mathbb N$, \begin{align*} d_Y(y_k,y)=d(y_k,y). \end{align*} Consequently, any eventual upper bound on the distances $d_Y(y_k,y)$ is exactly the same eventual upper bound on the ambient distances $d(y_k,y)$, and conversely.[/guided]

[step:Translate convergence in the subspace into convergence in the ambient space] Assume first that $(y_k)_{k=1}^{\infty}$ converges to $y$ in $(Y,d_Y)$. By the definition of convergence in a [metric space](/page/Metric%20Space), for every $\varepsilon>0$ there exists $N\in \mathbb N$ such that, for every $k\in \mathbb N$ with $k\ge N$, \begin{align*} d_Y(y_k,y)<\varepsilon. \end{align*} Using $d_Y(y_k,y)=d(y_k,y)$ for each such $k$, we obtain \begin{align*} d(y_k,y)<\varepsilon. \end{align*} Since this holds for every $\varepsilon>0$ with an eventual index $N$, the sequence $(y_k)_{k=1}^{\infty}$ converges to $y$ in $(X,d)$. [/step]

[step:Translate convergence in the ambient space into convergence in the subspace] Conversely, assume that $(y_k)_{k=1}^{\infty}$ converges to $y$ in $(X,d)$. By the definition of convergence in a metric space, for every $\varepsilon>0$ there exists $N\in \mathbb N$ such that, for every $k\in \mathbb N$ with $k\ge N$, \begin{align*} d(y_k,y)<\varepsilon. \end{align*} Using $d_Y(y_k,y)=d(y_k,y)$ for each such $k$, we obtain \begin{align*} d_Y(y_k,y)<\varepsilon. \end{align*} Therefore $(y_k)_{k=1}^{\infty}$ converges to $y$ in $(Y,d_Y)$. The two implications prove that convergence in the [metric subspace](/page/Metric%20Subspace) $(Y,d_Y)$ is equivalent to convergence in the ambient metric space $(X,d)$ for sequences contained in $Y$ with [limit point](/page/Limit%20Point) in $Y$. [/step]