finished #1-6

3.

Let $f: X \rightarrow X$ be continuous. Show that if $X = [0, 1]$, there is a point $x$ such that $f(x) = x$. We call $x$ a fixed point of $f$. What happens if $X$ equals $[0, 1)$ or $(0, 1)$?

Define $g: [0, 1] \rightarrow \mathbb{R}$ by $g(x) = f(x) - x$. By theorem 18.2 and 21.5, $g$ is continuous.

Note $g(0) = f(0) \geq 0$ and $g(1) = f(1) - 1 \leq 0$. Thus $0$ lies between $g(1)$ and $g(0)$, so by the intermediate value theorem there is some $c \in [0,1]$ such that $g(c) = 0$. Thus $f(c) - c = 0 \implies f(c) = c$ as required.

If $X = [0, 1)$ or $(0, 1)$, there is not necessarily a fixed point of $f$. Consider the function

$$ f(x) = \frac{1}{2} x + \frac{1}{2} $$

This function is continuous on $\mathbb{R}$ to $\mathbb{R}$, but it has no fixed point on $[0, 1)$ because $f(x) > x$ for all $x \in [0, 1)$.

4.

Let $X$ be an ordered set in the order topology. Show that if $X$ is connected, then $X$ is a linear continuum.

Suppose $X$ is an ordered set with the order topology, and it's connected.

First we show the least upper bound property.

Suppose $A$ is a nonempty subset of $X$ that is bounded above, but has no least upper bound. Since $A$ is bounded above, there is some $b \in X$ that is an upper bound for $A$.

Let $B$ be the set of all upper bounds for $A$, which is nonempty since it contains $b$. Now for any $x \in B$, there is some $x' \in B$ such that $x' < x$, because as we established, there is no least upper bound for $A$. Then $(x', \infty)$ is a neighborhood of $x$ contained in $B$. So $B$ is open.

Now consider $X - B$.

(i) If $X - B$ is empty, then $A \subset B$, so every $a \in A$ is an upper bound on $A$. By the simple order relation, this means that $A$ can only have one element (otherwise some element of $A$ is outside $B$). But then that one element is indeed the least upper bound for $A$, giving a contradiction.

(ii) If $X - B$ is nonempty, consider any $x \in X - B$. By definition $x$ is not an upper bound of $A$. In other words, there is some $a \in A$, such that $x < a$. And further, any other $x' < a$ is not an upper bound of $A$, and so it is not in $B$. Thus the set $(-\infty, a)$ is an open neighborhood of $x$ in $X - B$.

So $X - B$ is nonempty and open. But now we have a separation on $X$, which contradicts the fact that $X$ is connected.

Thus any nonempty subset of $X$ that is bounded above has a least upper bound.

Now we show the "in-between" property.

Suppose $x < y$ and there is no $w$ in $X$ such that $x < w < y$.

Then $(-\infty, y)$ and $(x, \infty)$ are disjoint, nonempty, and open in the order topology, so they constitute a separation of $X$, which contradicts the fact that it's connected.

Thus for any $x < y$ in $X$, there is some $w \in X$ such that $x < w < y$.