Let $k$ be an ordered field of cofinality $cf(k)$ whose Cauchy $cf(k)$-sequences are convergent.$^{(1)}$

Let $\mathcal{R}(k)$ be its real closure. As an algebraic extension of $k$, it has the same cofinality.$^{(2)}$

I wonder if $\mathcal{R}(k)$ has the same Cauchy completeness property as $k$, and if so, if this is true of any algebraic (ordered field) extension of $k$?

$^{(1)}$A map $u: cf(k) \rightarrow k$ is Cauchy if $\forall \varepsilon >_k 0(\exists \alpha \in cf(k)(\forall \beta,\gamma > \alpha(|u(\beta) - u(\gamma)| <_k \varepsilon)))$. It is convergent to $l \in k$ if $\forall \varepsilon >_k 0(\exists \alpha \in cf(k)(\forall \beta > \alpha(|u(\beta) -l| <_k \varepsilon)))$.

$^{(2)}$ If $(K,\rho)$ is an extension of $k$ then an induction on $d$ yields: $\forall d \in \mathbb{N}^*(\forall x \in K(\exists P \in \rho(k)_d[X](P(x) \leq 1 \longrightarrow [x;+\infty[ \cap \rho(k) \neq \varnothing)))$ so if $K / k$ is algebraic, $\rho(k)$ is cofinal. Therefore, $cf(K) \leq cf(k)$.

Now, given $E,E'$ cofinal in $k,K$ of order type $cf(k),cf(K)$, the range of the map $f:\alpha \in cf(K) \mapsto \min(\{\rho(x) \in \rho(E) \ | \ \rho(x) > E'^{\alpha}\})$ is cofinal in $\rho(E)$ so $|f(cf(K))| = cf(k)$, and $cf(K) \geq cf(k)$.