§ Closed graph theorem

• the graph of a function from a banach space to another banach space is a closed subset iff the function is continuous.
• Formally, given $f:X \to Y$, the set $G \equiv { (x, f(x)) }$ is closed in $X \times Y$ iff $f$ is continuous.

§ Proof: Continuous implies closed

• We must show that every limit point of the graph G is in G.

• Let $(p, q)$ be a limit point. Since everything in metric spaces is equivalent to the sequential definition, this means that $(p, q) = \lim (x_i, f(x_i))$.

• Limits in product spaces are computed pointwise, so $(p, q) = (\lim x_i, \lim f(x_i))$

• Thus, $p = lim xi$ from above. Now we calculate:

• $q = \lim f(x_i) = f (\lim xi) = f(p)$ where we use the continuity of $f$ to push the limit inside.
• Thus, $(p, q) = (p, f(p))$ which is an element of $G$.
• So an arbitrary limit point $(p, q) \in G$ is an element of $G$, and thus G is closed. Qed.

§ Proof: closed implies continuous

• Suppose G as defined above is a closed set. We must show that f is continuous, ie, $f$ preserves limits.

• Let $x_i$ be a sequence. We must show that $f(\lim x_i) = \lim f (x_i)$.

• Consider $(x_i, f(x_i))$ as a sequence in $G$. Let the limit of this sequence be $(p, q)$. Since G is closed, $(p, q)$ in G. By defn of $G$, $q = f(p)$.

• Now we compute that

• But since $q = f(p)$ (by defn of $G$), we have that $lim f(x_i) = q = f(p) = f(\lim x_i)$ which proves continuity.

§ Where this fails: topological spaces

• The graph of the identity function $I : X \to X$, $\Delta \equiv \{ (x, x) \in X \} \subset X \times X$is closed iff space is hausdorff.
• We call the set $\Delta$ the diagonal.

§ Closed implies Hausdorff

• Suppose the diagonal set $\Delta$ is closed.
• To show that the space is hausdorff, take two points $p, q \in X$.
• We must create two open sets $O_p, O_q$ which are disjoint, such that $p \in O_p$ and $q \in O_q$.
• Since $(p, q)$ is off the diagonal ( $p \neq q$), it belongs to the open set $X \times X - \Delta$.
• Thus, there must be some basic open set $O_p \times O_q$ such that $(p, q) \in O_p \times O_q$.
• Furthermore, the set $O_p \times O_q$ misses the diagonal.
• Recall that the product topology is generated by basic open sets, so we know that this $O_p, O_q$are open in $p$.
• We must have that $O_p$ is disjoint from $O_q$, since the set $O_p \times O_q$ misses the diagonal.
• Otherwise, we would have a $c \in O_p, O_q$ ( $c$ for contradiction), or that $(c, c) \in O_p \times O_q$, contradicting the fact that it misses the diagonal.

§ Hausdorff implies closed

• Suppose space is hausdorff. Now we need to show that the diagonal is closed.
• That is, we need to show that every limit point is included in the set.
• Consider some point $(p, q) \not \in S$. We will show that such a point cannot be a limit point.
• Since $(p, q) \not \in S$, we have that $p \neq q$.
• Since the space $X$ is hausdorff, there will be open sets $O_p, O_q$ such that they contain $p, q$respectively but are disjoint.
• This means that $O_p \times O_q$ is an open set in $X \times X$ which contains $p, q$, but is disjoint from $\Delta$. This means that $(p, q)$ is not a limit point.
• Thus, no point $(p, q)$ that is outside of $\Delta$ is a limit point, which means that $\Delta$contains all its limit points, and is thus closed.