# Math4201 Topology I (Lecture 10)

## Continuity

### Continuous functions

Let $X,Y$ be topological spaces and $f:X\to Y$. For any $x\in X$ and any open neighborhood $V$ of $f(x)$ in $Y$, $f^{-1}(V)$ contains an open neighborhood of $x$ in $X$.

#### Lemma for continuous functions

Let $f:X\to Y$ be a function, then:

1. $A\subseteq Y$: $f^{-1}(A^c) = (f^{-1}(A))^c$.
2. $\{A_\alpha\}_{\alpha\in I}\subseteq Y$: $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.
3. $\{A_\alpha\}_{\alpha\in I}\subseteq Y$: $f^{-1}(\bigcap_{\alpha\in I} A_\alpha) = \bigcap_{\alpha\in I} f^{-1}(A_\alpha)$.

<details>
<summary>Proof</summary>

1. By definition of continuous functions, $\forall V$ open in $Y$, $f^{-1}(V)$ is open in $X$.

2. It is sufficient to shoa that $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$ if and only if $x\in \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.

This condition holds if and only if $\exists \alpha\in I$ such that $f(x)\in A_\alpha$.

Which is equivalent to $\exists \alpha\in I$ such that $x\in f^{-1}(A_\alpha)$.

So $x\in f^{-1}(\bigcup_{\alpha\in I} A_\alpha)$

In particular, $f^{-1}(\bigcup_{\alpha\in I} A_\alpha) = \bigcup_{\alpha\in I} f^{-1}(A_\alpha)$.

3. Similar to 2 but use forall.

</details>

#### Properties of continuous functions

A function $f:X\to Y$ is continuous if and only if:

1. $f^{-1}(V)$ is open in $X$ for any open set $V\subset Y$.
2. $f$ is continuous at any point $x\in X$.
3. $f^{-1}(C)$ is closed in $X$ for any closed set $C\subset Y$.
4. Assume $\mathcal{B}$ is a basis for $Y$, then $f^{-1}(\mathcal{B})$ is open in $X$ for any $B\in \mathcal{B}$.
5. For any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.

<details>
<summary>Proof</summary>

**Showing $1\iff 3$**:

> Use the lemma for continuous functions (1)

**Showing $1\iff 4$**:

$1 \implies 4$: 

Because any $B\in \mathcal{B}$ is open in $Y$, so $f^{-1}(B)$ is open in $X$.

$4 \implies 1$:

Let $V\subset Y$ be an open set. Then there are basis elements $\{B_\alpha\}_{\alpha\in I}$ such that $V=\bigcup_{\alpha\in I} B_\alpha$.

So $f^{-1}(V) = f^{-1}(\bigcup_{\alpha\in I} B_\alpha) = \bigcup_{\alpha\in I} f^{-1}(B_\alpha)$ (by lemma (2)) is a union of open sets, so $f^{-1}(V)$ is open in $X$.

**Showing $1\implies 5$**:

Take $A\subseteq X$ and $x\in \overline{A}$. It suffices to show $f(x)$ is an element of the closure of $f(A)$. This is equivalent to say that any open neighborhood $V$ of $f(x)$ intersects $f(A)$ has a non-trivial intersection with $f(A)$.

For any such $V$, 1 implies that $f^{-1}(V)$ is open in $X$. Moreover, $x\in f^{-1}(V)$ because $f(x)\in V$.

This means that $f^{-1}(V)$ is an open neighborhood of $x$. Since $x\in \overline{A}$, we have $f^{-1}(V)\cap A\neq \emptyset$ and contains a point $x'\in X$.

So $x'\in f^{-1}(V)\cap A$, this implies that $f(x')\in V$ and $f(x')\in f(A)$, so $f(x')\in V\cap f(A)$.

> [!NOTE]
>
> This verifies our claim. Proof of $5\implies 1$ is similar and left as an exercise.

</details>

<details>
<summary>Example of property 5</summary>

Let $X=(0,1)\cup (1,2)$ and $Y=\mathbb{R}$ equipped with the subspace topology induced by the standard topology on $\mathbb{R}$.

Let $f:X\to Y$ be the inclusion map, $f(x)=x$ for all $x\in X$. This is continuous.

Let $A=(0,1)\cup (1,2)$. Then $\overline{A}=A$. So $f(\overline{A})=f(A)=(0,1)\cup (1,2)$.

However, $\overline{f(A)}=\overline{(0,1)\cup (1,2)}=[0,2]$.

So $f(\overline{A})\subsetneq \overline{f(A)}$.

</details>

#### Definition of homeomorphism

A **homeomorphism** $f:X\to Y$ is a continuous map of topological spaces that is a bijection and $f^{-1}:Y\to X$ is also continuous.

<details>
<summary>Example of homeomorphism</summary>

Let $X=\mathbb{R}$ and $Y=\mathbb{R}+$ with standard topology.

$f:\mathbb{R}\to \mathbb{R}^+$ be defined by $f(x)=e^x$ is continuous and bijective.

$f^{-1}:\mathbb{R}^+\to \mathbb{R}$ be defined by $f^{-1}(y)=\ln(y)$ is continuous and homeomorphism.

</details>

### Epsilon delta definition of continuity

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function where we use the standard topology on $\mathbb{R}$.

Then [property 4](#properties-of-continuous-functions) implies that for any open interval $(a,b)\in \mathbb{R}$, $f^{-1}((a,b))$ is open in $\mathbb{R}$.

Now take an arbitrary $x\in \mathbb{R}$ and $\epsilon > 0$. In particular $f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$ is an open set containing $x$.

In particular, there is an open interval (by the standard topology on $\mathbb{R}$) $(c,d)$ such that $x\in (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$.

Let $\delta = \min\{x-c, d-x\}$. Then $(x-\delta, x+\delta)\subseteq (c,d)\subseteq f^{-1}((f(x)-\epsilon, f(x)+\epsilon))$.

This says that if $|y-x| < \delta$, then $|f(y)-f(x)| < \epsilon$.