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Migrate to KaTeX

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Nathan Nguyen
2025-02-07 21:38:28 -06:00
parent 41af035e50
commit c7c46c56ee
27 changed files with 369 additions and 391 deletions
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docs/math/real-analysis/5-continuity.md
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@@ -2,10 +2,10 @@
## Section 5.1 - Continuous Functions
**Definition**. Let $A \subseteq \mathbb{R}$, and $f: A \rightarrow \mathbb{R}$. Then, if $a \in A$, $f$ is *continuous at $a$* if, given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in A$,
**Definition**. Let $A \subseteq \mathbb{R}$, and $f: A \rightarrow \mathbb{R}$. Then, if $a \in A$, $f$ is *continuous at $a$* if, given any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in A$,
$$
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon
$$
Note that if $a$ is an *isolated point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
@@ -20,10 +20,10 @@ Note that a function cannot be continuous at a point outside of its domain, even
---
**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
$$
d_S(x, a) < \delta \Rightarrow d_T(f(x), f(a)) < \epsilon
d_S(x, a) < \delta \Rightarrow d_T(f(x), f(a)) < \varepsilon
$$
**Theorem**. A function $f: S \rightarrow T$ is continuous at a point $a \in A$ if and only if given some neighborhood $V(f(a)) \in B$, there exists some $U(a) \in A$ such that $f(U) \subseteq V$.
@@ -63,10 +63,10 @@ $$
---
**Definition**. Let $A \subseteq R$. Then, a function $f: A \rightarrow \mathbb{R}$ is *uniformly continuous* if given any $\epsilon > 0$, there exists some $\delta > 0$ depending only on $\epsilon$ such that for any $x, y \in A$,
**Definition**. Let $A \subseteq R$. Then, a function $f: A \rightarrow \mathbb{R}$ is *uniformly continuous* if given any $\varepsilon > 0$, there exists some $\delta > 0$ depending only on $\varepsilon$ such that for any $x, y \in A$,
$$
|x - y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon
|x - y| < \delta \Rightarrow |f(x) - f(y)| < \varepsilon
$$
Note that if $f$ is uniformly continuous, it must be continuous on $A$.