The preimage of a closed/open set is closed/open under a continuous function. The same is NOT generally true about images.
The inverse of a continuous bijection is not generally continuous.

Limits of Functions

For a function $f : E \rightarrow Y$, where $E\sube X$ and $X, Y$ are metric spaces, and a limit point $p$ of $E$, we say that $q \in Y$ is the limit of $f$ as $x$ approaches $p$ if the following holds:

$$ \forall \varepsilon > 0 , \exists \delta >0 : 0<d_X(x, p) < \delta \implies d_Y(f(x), q) < \varepsilon \hspace{10pt} \forall x\in E $$

Convergent Sequences in Range Equivalence. A function $f: E \rightarrow Y$ limits to $q$ as $x$ approaches $p$ iff for every sequence $\{x_n\}$ that converges to $p$, the sequence $\{f(x_n)\}$ converges to $q$.

$$ \lim _{x\rightarrow p} f(x)= q \iff [x_n \rightarrow p \implies f(x_n) \rightarrow q] $$

Proof </aside>

Limits of Functions are Unique.

Proof </aside>

Limit Laws for Complex-Valued Functions. Suppose that $X$ is a metric space, $E \sube X$, and $p$ is a limit point of $E$. Then, for functions $f, g : E \rightarrow \mathbb{C}$ whose limits at $p$ are $a$ and $b$, respectively, we have the following:

$$ \lim _{x \rightarrow p} (f+g)(x) = a + b \hspace{157pt} \\ \lim _{x \rightarrow p } (fg)(x) = ab \hspace{182pt} \\

\lim _{x \rightarrow p } (\frac{f}{g}) (x) = \frac{a}{b} \hspace{15pt} \text{$b\neq0$ and $\exists B_r(p) : \forall x \in B_r(p), g(x) \neq 0$} $$

</aside>

Continuity

Consider a function $f : X \rightarrow Y$, where $X, Y$ are metric spaces. We say $f$ is continuous at $p \in X$ if the following holds:

$$ \forall \varepsilon >0 , \exists \delta >0 : d_X(x, p) < \delta \implies d_Y(f(x), f(p)) < \varepsilon $$

Moreover, if $f$ is continuous at every point in its domain, we say $f$ is simply continuous.

Alternative. If $p$ is a limit point, then $f$ is continuous at $p$ iff $\lim _{x \rightarrow p} f(x) = f(p)$. Otherwise, $f$ is continuous at $p$ vacuously.

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Limit of Evaluation Sequence is Evaluation of Limit. If $f$ is continuous at a limit point $p$, we have that for any sequence $x_n \rightarrow p$:

$$ \lim_{n\rightarrow \infty} f(x_n) = f(\lim_{n\rightarrow \infty} x_n) $$

Proof </aside>

Topological Characterization. A function $f : X \rightarrow Y$ is continuous iff the preimage of any open subset of $Y$ is an open subset of $X$.

$$ f \text{ is continuous} \iff \text{For all open $V \sube Y$, $f^{-1}(V) \sube X$ is open} $$

Proof </aside>

Continuity Preserved under Composition. Consider metric spaces $X, Y, Z$ and functions $f: E \sube X \rightarrow Y$ and $g : D \sube Y \rightarrow Z$ such that $f(E) \sube D$. Then if $f$ is continuous at $p$ and $g$ is continuous at $f(p)$, then $g \circ f$ is continuous at $p$.

Proof </aside>