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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.

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3 Answers

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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 $$ d(x,0)\leq d(x_1,0)+r/2<r/2+r/2=r. $$

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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.

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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, $d(x,0)<r$

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