[proofplan] We establish a collection of equivalent and sufficient conditions for closedness of a subspace of a Hilbert space. The equivalences (i)--(iv) connect closedness to sequential closedness, completeness, and the double orthogonal complement. The sufficient conditions cover finite-dimensionality, kernels of bounded operators, and preimages of closed sets. Each direction is proved independently using standard metric space and Hilbert space tools. [/proofplan] [step:Prove the sequential characterisation: (i) $\Leftrightarrow$ (ii)] This is the general topological fact for metric spaces: a subset of a metric space is closed if and only if it contains all its limit points, which is equivalent to containing the limits of all convergent sequences. [/step] [step:Prove the completeness characterisation: (i) $\Leftrightarrow$ (iii)] [claim:Closedness equals completeness for subspaces] A subspace $M$ of a [Hilbert space](/page/Hilbert%20Space) $H$ is closed if and only if $M$ is complete as a metric space with the inherited metric. [/claim] [proof] ($\Rightarrow$) If $M$ is closed and $\{x_n\} \subset M$ is Cauchy, then $\{x_n\}$ is Cauchy in $H$ (same metric), so $x_n \to x$ for some $x \in H$ by completeness of $H$. Since $M$ is closed, $x \in M$. ($\Leftarrow$) If $M$ is complete and $\{x_n\} \subset M$ converges to $x \in H$, then $\{x_n\}$ is Cauchy in $M$, so it converges to some $y \in M$. By uniqueness of limits, $x = y \in M$. [/proof] [/step] [step:Prove the double orthogonal complement characterisation: (i) $\Leftrightarrow$ (iv)] [claim:Closedness equals $M = M^{\perp\perp}$] A subspace $M$ of a Hilbert space $H$ is closed if and only if $M = M^{\perp\perp}$. [/claim] [proof] ($\Rightarrow$) This is a consequence of the [Orthogonal Decomposition Theorem](/theorems/241): when $M$ is closed, $H = M \oplus M^\perp$, and applying the decomposition again to $M^\perp$ gives $M^{\perp\perp} = M$. ($\Leftarrow$) The orthogonal complement $M^\perp$ is always closed (as the intersection of kernels of the continuous linear functionals $x \mapsto (x, m)_H$ for $m \in M$). Therefore $M^{\perp\perp} = (M^\perp)^\perp$ is also closed. If $M = M^{\perp\perp}$, then $M$ is closed. [/proof] [/step] [step:Prove finite-dimensional subspaces are closed: (v)] [claim:Finite-dimensional subspaces are closed] Every finite-dimensional subspace of a Hilbert space is closed. [/claim] [proof] Let $M$ have an orthonormal basis $\{e_1, \dots, e_d\}$ (obtained by Gram--Schmidt). Let $\{x_n\} \subset M$ converge to $x \in H$. Write $x_n = \sum_{k=1}^d \alpha_k^{(n)} e_k$. For each $k$: \begin{align*} \alpha_k^{(n)} = (x_n, e_k)_H \to (x, e_k)_H =: \alpha_k. \end{align*} By continuity of the inner product, the coefficients converge, so $x_n \to \sum_{k=1}^d \alpha_k e_k \in M$, and by uniqueness of limits, $x \in M$. [/proof] [/step] [step:Prove kernels of bounded operators are closed: (vi) and (vii)] If $T: H \to Y$ is a bounded linear operator into a normed space $Y$, then $\ker T = T^{-1}(\{0\})$ is the preimage of the closed set $\{0\}$ under the continuous map $T$, hence closed. More generally, for any bounded linear operator $T: H \to Y$ and any closed subset $C \subseteq Y$, the preimage $T^{-1}(C)$ is closed because continuous preimages of closed sets are closed. Since bounded linear operators between normed spaces are continuous, this applies to any such $T$. [/step]