# Sum of closed subspaces of normed linear space

Problem Suppose $R$ is a normed linear space, then show that:

If $M$ is closed subspace of $R$ and $N$ a finite dimensional subspace of $R$, then the set
$$M+N=\{ z : z = x + y , x \in M , y \in N \}$$ is a closed subspace of $R$.

What I've done I know being finite dimensional makes $N$ closed. Also I can disjointize $M$ and $N$ so that I have a direct sum instead of sum. But I don't know if the direct sum of closed subspaces are closed. I try taking a convergent sequence but i cant control the limit point. I even tried induction but cant show the 1-dimensional case. Could you please help me?

Look at the quotient space $X/M$. Because $M$ is known to be closed, then $Y=X/M$ is a normed space under the quotient norm. Let $\phi : X\rightarrow X/M$ be the quotient map. Then $\phi$ is continuous because $\|\phi(x)\|_{X/M}\le \|x\|_{X}$. Furthermore, $\phi(M+N)$ is finite-dimensional and, hence, closed in $X/M$. Therefore $M+N=\phi^{-1}(\phi(M+N))$ is closed.

• Could you please explain why $\phi(M+N)$ is finite-dimensional? Feb 14, 2019 at 17:50
• @MadhurPanwar : $\phi(x)=[x]$ where $[x]=\{ x+m : m \in M \}$. If $x\in M+N$, then $x=m+n$ and $[x]=[n]+[m]=[n]$ because $[m]=[0]$. Feb 14, 2019 at 20:22
• @DisintegratingByParts Why is $M+N = \phi^{-1}(\phi(M+N))$? $\phi$ is clearly not injective. Can you argue that for every $[x] \in \phi(M+N)$ there is a $n\in N$ s.t. $[x]=[n]$ (is that correct?), so $x\in M+N$? May 1, 2019 at 10:48
• @user3766553 I think what the answerer have meant to say is that $M + N = \phi^{-1} \left ( \phi (N) \right ),$ which can be shown quite easily. Now observe that $\phi (N)$ is finite dimensional since $N$ is so and hence $\phi (N)$ is a closed subspace of $X/M.$ Since $\phi$ is continuous it follows that $M + N$ is a closed subspace of $X,$ as required. Oct 30, 2020 at 6:20
• $\phi^{-1}(K)$ is the inverse image of the closed set $K$, which is closed due to continuity of $\phi$. Hence, taking $K=\phi(N)=\phi(M+N)$, the desired result arises. Mar 18, 2021 at 18:55

Word to the wise: the sum of closed subspaces is not necessarily closed, sadly. If one is finite-dimensional, however, this is true.

To prove this: suppose $N \cap M = \{0\}$. Since $N$ is finite-dimensional, we have that $S_N = \{v \in N \mid |v| = 1\}$ is compact. So, because $S_N$ and $M$ are disjoint and $v \mapsto d(v, M) = \inf\{|v - m \mid m \in M\}$ is continuous, $$\alpha(N, M) := \inf_{v \in S_N} d(v,M) > 0$$ Now you want to show that the projection $P : N \oplus M \to N$ is bounded. We compute $$|P| = \sup_{v = m + n,m \in M, n \in N, v \neq 0} \frac{|n|}{|m + n|} = \sup_{|n| = 1, m \in M} \frac{1}{|n - m|} = \frac{1}{\alpha(N, M)} < \infty$$ The boundedness of $P$ implies that $M \oplus N$ is closed; to see this, note that $z_n \in M \oplus N, z_n \to z$ implies $P z_n, (I -P) z_n$ both converge.

• This proof is a little longer, but I have found that this notion of 'angles between subspaces' is not well-advertised, so I gave this method. Jan 15, 2014 at 19:08
• so does the projection being bounded implies closedness of $M+N$? Also could please explain what $|P|$ is. Jan 15, 2014 at 19:27
• @iamvegan Yes, the boundedness of the projection implies that $M \oplus N$ is closed. You should try to prove this as an exercise. Also, $|P|$ is supposed to be the operator norm of $P$ induced by the norm on $M + N$. Jan 15, 2014 at 19:51