Prop: If $f: X \to Y$ is continuous, and $K \In X$ is compact, then $f(K)$ is compact.
Proof: any open cover of $f(K)$ can be pulled back to be an open cover of $K$, then we can pick a finite subcover in the domain, and the corresponding cover in the target forms a cover of $f(K)$.

We can also prove using sequential compactness. To see any sequence in $f(K)$ subconverge, we can lift that sequence back to $K$, find a convergent subsequence, and push them back to $f(K)$.

Lemma: if $f: X \to Y$ is continuous, then for any $E \In X$, $f|_E: E \to Y$ is continuous.

Lemma: if $f: X \to Y$ is continuous, then $f: X \to f(X)$ is continuous.

Prop: If $f: X \to Y$ is continuous, and $K \In X$ is connected, then $f(K)$ is connected.
Pf: first, note that map $f: K \to f(K)$ is also continuous. If $f(K)$ is the disjoint union of two non-empty open subsets, $f(K) = U \cap V$, then $K = f^{-1}(U) \cap f^{-1}(V)$ the disjoint union of two non-empty open subsets of $K$.

Intermediate value theorem: if $[a,b] \In \R$, and $f: \R \to \R$ is continuous, then $f([a,b])$ is also a closed interval.
Proof: since $[a,b]$ is compact, hence $f([a,b])$ is compact, hence closed. Since $[a,b]$ is connected, hence $f([a,b])$ is connected, hence an interval, a closed interval.

We say a function $f: X \to Y$ is uniformly continuous, if for any $\epsilon >0$, there exists $\delta > 0$, such that for any pair $x_1, x_2 \in X$ with $d(x_1, x_2)<\delta$, we have $d(f(x_1), f(x_2))< \epsilon$.

For example, the function $f: (0, 1) \to \R$ $f(x) =1 /x$ is continuous but not uniformly continuous.

Theorem: if $f: X \to Y$ is continuous, and $X$ is compact, then $f$ is uniformly continuous.

Proof: Fix $\epsilon>0$. For each $x \in X$, let $r_x > 0$ be small enough such that $f(B_{2r_x}(x)) \In B_{\epsilon/2}(f(x))$. Let $B_x = B_{r_x}(x)$. The, $\{B_x: x \in X\}$ forms an open cover of $X$. Pass to finite subcover, $X = \cup_{i=1}^N B_{x_i}$. Then let $\delta = \min \{r_{x_i}\}$. Then, suppose $x$ and $x'$ has distance less than $\delta$, then $x$ is contained in some $B_{r_{x_i}}(x_i)$, and $x' \in B_{2 r_{x_i}}(x_i)$, thus $f(x), f(x') \in B_{\epsilon/2}(f(x_i))$, thus $f(x)$ and $f(x')$ has distance less than $\epsilon$.

Now we will leave the safe world of continuous functions. We consider more subtle cases of maps.

Def: Let $f: X \to Y$ be any map, and let $x \in X$ be a point, we say $f$ is continuous at $x$, if for any $\epsilon>0$, there exists $\delta>0$, such that $f(B_\delta(x)) \In B_\epsilon(f(x))$.

Def: Let $E \In X$ and $f: E \to Y$. Suppose $x \in \bar E$. We say $\lim_{p \to x} f(p) = y$ if for any convergent sequence $p_n \to x$ with $p_n \in E$, we have $f(p_n) \to y$.

note that $x$ may not be in $E$.

Prop: $f: X \to Y$ is continuous at $x \in X$, if and only if $\lim_{p \to x} f(p) = f(x)$.

If $f: (a,b) \to \R$ is a function, and $f$ is not continuous at some $x \in (a,b)$, then

• if $\lim_{t \to x-} f(t)$ and $\lim_{t \to x+} f(t)$ both exists, but does not equal to $f(x)$, we say this is a simple discontinuity, or first kind discontinuity
• otherwise, it is called a second kind.