Locally compact space isometric to an open subset of its completion

Question

Let $(X,d)$ be a locally compact metric space.

(For every $x$ in $X$, there exists a ball centred in $x$ whose closure is compact.)Let $(Y, δ)$ be its completion. I search to prove that $(X,d)$ is isometric to an open subset of $(Y,δ)$.

I tried to prove that $φ(Χ)$ is open , where $φ$ is the isometric injection of the proposition of the existence of a completion. And to use that the range of a compact is compact.

Answer 1

Your idea is good: $\varphi(X)$ is indeed open in $Y$. Let $r>0$ be such that $B_r^X(x)(=\{x'\in X\,|\,d(x,x')\leqslant r\})$ is a compact neighborood of $x$ (in $X$). Then $\varphi\left(B_r^X(x)\right)$ is a compact subset of $Y$ and therefore it cannot contain any element $y\in Y\setminus\varphi(X)$, because $y$ would be the limit of a sequence $(x_n)_{n\in\mathbb N}$ of elements of $U$ and, since $U$ is compact, that sequence has a subsequence $(x_{n_k})_{k\in\mathbb N}$ which converges, at the same time, to $y(\notin\varphi(X))$ and to an element of $B_r^X(x)(\subset X)$. By the same argument, $B_r^X(x)=B_r^Y(x)$.

So, $x\in B_r^Y(x)$ and $B_r^Y(x)\subset\varphi(X)$. Since this takes place for each $x\in X$, $\varphi(X)$ is an open subset of $Y$.

Answer 2

With some indications of your proof and some other helps, I tried to write it by own. I will be thankful to have some mathematical, formal or English critics.

Let $y$ be an element of $φ(x)$, there exists $x$ in $X$ such that $y=φ (x) $. Since $x$ lies in $X$, which is locally compact by assumption, there exists $ε>0$ such that $B_F(x,ε)$ is compact.

Take first $z\in B(y,ε)$, we want to show that $z$ belongs to $φ(Χ)$.

To that purpose, let's take $z_n$ a sequence of $φ(Χ)$ which converge to $z$. We are allowed to do it because $φ(Χ)$ is dense in $(Y,δ)$.

In particular, after a certain rank $z_n \in B(y,ε) \cap φ(Χ)$. We can also denote $x_n$ the sequence such that $z_n=f(x_n)$.

Moreover $φ$ is an isometric induction, it follows that $d(x_n,x)<ε$. It believes true for the large inequality. Consequently , $x_n$ is a sequence of our compact $B_F(x,ε)$. We extract a convergent subsequence $x_{ψ(k)}$ which converge to $x_\infty $.

Then we have the following inequality $$ δ(φ(x),z) \leq δ(φ(x_\infty), φ(x_{ψ(n)}) + δ(z_{ψ(n)}),z) , $$ where both terms tends to zero. That's why $φ(x_\infty) =z$.

As expected, we obtained that $z$ belongs to $φ(Χ)$. We can conclude that $B(y,ε)\subset φ(Χ)$ which proves the required claim.