Prove a subspace of a Banach space is closed

Let $X$ be a Banach space, $M$ and $N$ are two closed linear subspace of $X$. $N$ is finite dimensional.Prove $M+N$ is a closed subspace of $X$.

It's trival to check that $M+N$ is a space. Then what I need to do is to prove $M+N$ is closed. First I get $M\cap N\subset N$ is a finite dimensional space and the basis of $M\cap N$ is {$\alpha_1,\cdots,\alpha_r$}. The basis of $N$ is {$\alpha_1,\cdots,\alpha_r,\alpha_{r+1},\cdots,\alpha_n$}. Then I get $M+N=M+<\alpha_{r+1},\cdots,\alpha_n>$.

So I can always assume that $M\cap N=\{0\}$,$M+N=M\oplus N$.

Then I want to prove the limit of the Chauchy sequence {$x_n+y_n$},$x_n\in M,y_n\in N$ is in $M\oplus N$, I try to prove {$y_n$} is bounded in $N$ but failed. If {$y_n$} must be bounded or not necessary?

• $M$ should be assumed closed, otherwise $M + N$ need not be closed. – Daniel Fischer Apr 20 '15 at 13:23
• I fogot it, thank you. – yahoo Apr 20 '15 at 13:45

First of all you need $M$ to be a closed subspace. To see this consider the space $M=P[0,1]$ of all polynomials in $X=C[0,1]$. Now, let $N = P_2[0,1]$, the space of all polynomials of degree less than or equal to $2$. Then $N$ is finite dimensional but $M+N = M$, which is not closed in $X$.
Now suppose that $M$ is closed in $X$ and $N$ is finite dimensional. Consider $X/M$ and the quotient map $q:X \to X/M$. Since $\dim q(N) \leq \dim N < \infty$, it follows that $q(N)$ is closed in $X/M$. Since $q$ is continuous, so, $q^{-1}(q(N))$ is closed in $X$; but $q^{-1}(q(N)) = M+N$. This completes the proof.