Week 6 Discussion Notes

Solution.

Assume that $A \cup B = U \cup V$, where $U, V$ are disjoint, non-empty, open sets in $A \cup B$. By the properties of sub-metric spaces, there exist open sets $U_X, V_X \subseteq X$ in $X$ such that $U = \p{A \cup B} \cap U_X$ and $V = \p{A \cup B} \cap V_X$.

Claim: $A \subseteq U$ or $A \subseteq V$.

Suppose otherwise, and that $A \cap U \neq \emptyset$ and $A \cap V \neq \emptyset$. Notice that

i.e., $A \cap U$ is open in $A$ and by symmetry, $A \cap V$ is open in $A$ also. But $A \cup B = U \cup V$, so

But $A \cap U$ and $A \cap V$ are disjoint, non-empty, and open sets in $A$, which implies that $A$ is disconnected, a contradiction.

Thus, $A \subseteq U$ or $A \subseteq V$. Without loss of generality, we may assume that $A \subseteq U$. In the other case, we can just replace $U$ with $V$ in the following argument.

Because $A \cap B \neq \emptyset$, this means that $B \cap U \neq \emptyset$. Since $B$ is connected, this means that $B \subseteq U$ as well by using the same argument in the claim, which means that $A \cup B \subseteq U$, so $V = \emptyset$, a contradiction. Thus, no such $U$ and $V$ existed to begin with, i.e., $A \cup B$ is connected.

Solution.

Assume otherwise, and that $f$ is not constant. Then $E^\comp \neq \emptyset$: since $E \neq \emptyset$, there exists $x \in X$ such that $f\p{x} = y$, but $f$ is not constant, so there exists $x' \in X$ such that $f\p{x'} \neq y$, i.e., $x' \in E^\comp$.

Since $E$ was closed, we know that $E^\comp$ is open, and by definition,

But $E, E^\comp$ are disjoint (by definition), non-empty, and open, which implies that $X$ is disconnected, a contradiction.

Solution.

We'll show that $E$ is path-connected. Let $f, g \in E$, and define the line segment (or homotopy) between them via $\func{\gamma}{\br{0,1}}{E}$,

For any $t \in \br{0, 1}$, the function $\gamma_t\p{x}$ is continuous since it's a linear combination of the continuous functions $f$ and $g$, and

so $\gamma$ is well-defined (i.e., it actually gives me functions in $E$).

If $f = g$, then $\gamma$ is constant in $t$, so it's automatically continuous. If $f \neq g$, then let $\epsilon > 0$.

$C_{f,g}$ exists and is finite because $\abs{f - g}$ is a continuous function on the compact set $\br{0, 1}$. Moreover, $C_{f,g} \neq 0$ since we assumed that $f \neq g$. Thus, we can pick $\delta = \frac{\epsilon}{C_{f,g}}$. Then if $\abs{t - s} < \delta$, we get

so $\gamma$ is continuous in $t$. In other words, $\gamma$ is a valid path between $f$ and $g$, so $E$ is path-connected, hence connected.

Solution.

We can define $\func{F}{\R}{\Gamma}$ by $F\p{x} = \p{x, f\p{x}}$. By construction, $\Gamma = F\p{\R}$, and since $\R$ is connected, if we show that $F$ is continuous, we automatically know that $\Gamma$ is connected.

Let $\epsilon > 0$ and fix $x_0 \in \R$. Recall that

Since $f$ is continuous, there exists $\delta > 0$ such that if $\abs{x - y} < \delta$, then $\abs{f\p{x} - f\p{y}} < \frac{\epsilon}{\sqrt{2}}$. Thus, if $\abs{x - y} < \min\set{\delta, \frac{\epsilon}{\sqrt{2}}}$, then

so

so $F$ is continuous, which means $\Gamma$ is connected.