I am confused by a step in the Gamelin and Greene proof of the Baire Category Theorem.

Here is the start of the proof.

Theorem: Let $\{U_n\}_{n=1}^{\infty}$ be a sequence of dense open subsets of a complete metric space $X$. Then $\cap_{n=1}^{\infty}U_n$ is also dense in $X$.

Proof: Let $x \in X$ and let $\epsilon > 0$. It suffices to find $y \in B(x;\epsilon)$ that belongs to $\cap_{n=1}^{\infty} U_n$. Indeed, then every open ball in $X$ meets $\cap U_n$, so that $\cap U_n$ is dense in $X$.

Since $U_1$ is dense in X, there exists $y_1 \in U_1$ such that $d(x,y_1) < \epsilon$. Since $U_1$ is open, there exists $r_1 > 0$ such that $B(y_1;r_1) \subset U_1$.

... so far so good, I get it this far. They continue...

By shrinking $r_1$, we can arrange $r_1 < 1$, and $\overline{B(y_1;r_1)} \subset U_1 \cap B(x,\epsilon)$.

This is where I am confused. I don't like this fuzzy use of "shrink" here but get what they mean I think. I see that we could shrink $B(y_1;r_1)$ as they indicate but I don't see how the closure $\overline{B(y_1;r_1)}$ can be shrunk. Could there be an unusual metric space where the open ball is dense in $X$ or something similar that we just cant think of offhand which would make this impossible? How can we be sure this is true in general in a rigorous way?

Thanks for your help.


By the triangle inequality, for each $x$ and $r > 0$, the set

$$\{ y\in X : d(x,y) > r\}$$

is open: $B(y; d(x,y)-r)$ is contained in it. Therefore $\{ y \in X : d(x,y) \leqslant r\}$ is closed, and

$$\overline{B(x;r)} \subset \{ y \in X : d(x,y) \leqslant r\} \subset B(x; r+\varepsilon)$$

for every $\varepsilon > 0$.

So, the closure of a ball with radius $r$ is contained in every ball with larger radius (and the same centre, of course). Thus by shrinking the radius by any small margin, you obtain a ball whose closure is contained in the original.

  • $\begingroup$ I'm slow. Can you please explain why $\overline{B(x;r)} \subset \{y \in X : d(x,y) <= r\}$ follows from $\{y \in X : d(x,y) <= r\}$ being closed? $\endgroup$ – topoquestion Jan 9 '14 at 0:30
  • 2
    $\begingroup$ The closure is the smallest closed set containing the ball. $\{ y\in X : d(x,y) \leqslant r\}$ is a closed set containing $B(x;r)$, hence it contains the closure. While in general metric spaces the closed ball $K(x;r) = \{y\in X : d(x,y)\leqslant r\}$ is rarely the closure of $B(x;r)$, you always have $\overline{B(x;r)} \subset K(x;r)$. $\endgroup$ – Daniel Fischer Jan 9 '14 at 0:39
  • $\begingroup$ okay, I remember that from exercise 7 in the last chapter now. That was the key I was missing before I think. $\endgroup$ – topoquestion Jan 9 '14 at 0:57

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