Rudin's RCA The Open Mapping Theorem

Question

There is the definition which we need for the theorem:

This is the Baire's theorem:

If $X$ is a complete metric space, the intersection of every coutanble collection of dense open subsets of $X$ is dense in $X$ .

I don't understand how does the Baire theorem imply that

there is a nonempty open set $W$ in the closure of some $\Lambda(kU)$.

I also don't understand how does the linearity of $\Lambda$ shows that the last inequality holds if $\delta$ $=$ $\frac {\tau}{2k}$ ?

Any help would be appreciated.

Answer 1

Usually, the Baire theorem is written like this:

If $X$ is complete, then the intersection of a countable collection of dense open subsets of $X$ is dense in $X$.

But you can also write the Baire theorem like this:

If $X$ is complete, then the union of a countable collection of closed subsets of $X$ with empty interior has empty interior.

To see that the two statements are equivalent, simply observe that:

• A set is open iff its complement is closed.
• A set is dense iff its complement has empty interior.

In your proof, the $\overline{\Lambda(k U)}$'s are a countable collection of closed subsets of $Y$, and $Y$ is a complete metric space. The union of the $\overline{\Lambda(k U)}$'s is the whole of $Y$, and so, the union of the $\overline{\Lambda(k U)}$'s certainly cannot have empty interior. Therefore, at least one of the $\overline{\Lambda(k U)}$'s has non-empty interior - which is to say that some $\overline{\Lambda(k U)}$ contains a non-empty open set $W$.

---

Regarding your second question, we know that for all $y$ such that $\left\| y \right\| < \eta $, there exists a sequence $\{ x_i \}$ with $\left\| x_i \right\| < 2k $ such that $\Lambda x_i \to y$.

Said another way, for all $y$ such that $\left\| y \right\| < \eta $ and for all $\epsilon > 0$, there exists an $x$ with $\left\| x \right\| < 2k $ such that $\left\| \Lambda x - y \right\| < \epsilon $.

Let's now prove an alternative version of the above statement, with one of the $<$ symbols replaced by $\leq$ symbols. Let's prove that for all $y$ such that $\left\| y \right\| \leq \eta $ and for all $\epsilon > 0$, there exists an $x$ with $\left\| x \right\| < 2k $ such that $\left\| \Lambda x - y \right\| < \epsilon $.

To prove this, observe that there exists a $\tilde y$ with $\| \tilde y \| < \eta$ and $\| \tilde y - y \| < \tfrac 1 2 \epsilon$. (For example, take $\tilde y = (1 - \min(\tfrac 1 4 \epsilon / \| y \|, \tfrac 1 2) ) y$.) We know that there exists an $x$ with $\left\| x \right\| < 2k $ such that $\left\| \Lambda x - \tilde y \right\| < \tfrac 1 2 \epsilon $. So $\left\| \Lambda x - y \right\| \leq \left\| \Lambda x - \tilde y \right\| + \left\| \tilde y - y \right\| < \epsilon$.

Now consider a general $y$, and pick a $\delta > 0$. Fix an $\epsilon > 0$. Define $y' = (\eta / \left\| y \right\|) y $. We have $\left\| y' \right\| \leq \eta$. So the statement a couple of paragraphs earlier applies to $y'$. Applying this statement with $\epsilon ' = \epsilon \eta /\left\| y \right\|$, we find that there exists an $x'$ with $\left\| x' \right\| < 2k $ such that $\left\| \Lambda x' - y' \right\| < \epsilon ' = \epsilon \eta /\left\| y \right\| $.

Define $x = (\left\| y \right\| / \eta) x' $. We have $\left\| x \right\| \leq 2k \left\| y \right\| / \eta = \delta^{-1}\left\| y \right\| $. Furthermore, since $\Lambda$ is linear, we have $\Lambda x = (\left\| y \right\| / \eta) \Lambda x' $. Thus $\| \Lambda x - y \| = (\left\| y \right\| / \eta) \| \Lambda x' - y' \| < \epsilon $, which is what you wanted to prove.