# A question about the proof of the open mapping theorem in “Functional Analysis” by Rudin

In the proof of the open mapping theorem in "Functional Analysis" by Rudin, there is the following argument:

Let $X$ be a topological vector space in which its topology is induced by a complete invariant metric $d$. Define $V_n = \{ x \in X : d(x,0) < 2^{-n}r \}$ for $n=1,2,\cdots$, where $r>0$. Suppose $x_n \in V_n$. Since $x_1 + \cdots + x_n$ form a Cauchy sequence, it converges to some $x \in X$ with $d(x,0) < r$.

In the above, I cannot understand why $d(x,0)<r$. It seems to me that $d(x,0)\leq r$.

Any help would be appreciated.

-

Let $\delta=r/2-d(x_1,0)>0$. Then $$d(x,0)\leq d(x,x_1+\cdots+x_n)+d(x_1+\cdots+x_n,0)\leq d(x,x_1+\cdots+x_n) +\sum_1^nd(x_k,0)\leq d(x,x_1+\cdots+x_n)+d(x_1,0)+\sum_2^n2^{-k}r.$$ Taking limsup, we get $$d(x,0)\leq d(x_1,0)+r/2<r/2+r/2=r.$$

-

I prefer to be in a normed space instead of a metric space and write $\|x\|$ instead of $d(x,0)$ (personally, I find it easier to read and think about). Then you have that $x$ in $V_n$ are such that $\|x\| < r 2^{-n}$.

Note that $\sum_{k=1}^\infty \frac{1}{2^k} = 1$ and hence $\sum_{k=1}^\infty \|x_k\| < r \sum_{k=1}^\infty \frac{1}{2^k} = r$.

Hope this helps.

-

The limit of the summation goes to r and so you might be thinking that $d(x,0)\leq r$. But each $x_{n}$ is chosen such that $d(x_{n},0)< 2^{-n}r$ and so the limit of cauchy sequence will be less than $r$, i.e, $d(x,0)<r$

-