Examples

Simple topology - connectedness

Simple topology - connectedness

Three points: $X=\{A,B,C\}$, with two (nontrivial) open sets.
This is not a topology because the intersection $\{B\}$ is not open:

This is a topology, and it is connected:

Continuum - real number interval

Continuum - real number interval

The standard topology structure for real numbers is generated by open intervals $(a,b)$.

(“Generated”: it is the smallest topology structure containing all such intervals.)

In $n$D, the open sets are generated by open boxes $(a,b)\times (c,d)$ or (equivalently) open balls ${\mathbf{B}}_{\epsilon }(\vec{c})=\{\vec{x}\in {\mathbb{R}}^n\mid |\vec{x}-\vec{c}|<\epsilon \}$.

Open in S but not in R2

Open in $S$ but not in ${\mathbb{R}}^2$

Consider the subspace $S=[0,1{]}^2\subset {\mathbb{R}}^2$ equipped with the subspace topology inherited from the topology structure on ${\mathbb{R}}^2$:

An open set in $S$ may be given as the intersection with a ball in ${\mathbb{R}}^2$, but this set is not open in ${\mathbb{R}}^2$ because the top and right edges would be included. No open ball can be found around a point in this edge and contained in $S$.

Open-set continuity agrees with epsilon-delta continuity

Open-set continuity agrees with $\epsilon$-$\delta$ continuity

Assume that $f$ is continuous according to the topological definition.

1. Select any $x_0\in \mathbb{R}$. Let $\epsilon >0$ be given.
2. The interval $V=(f(x_0)-\epsilon ,f(x_0)+\epsilon )$ is open.
3. By the topological definition of continuity, the preimage $f^{-1}(V)$ is open. It contains $x_0$.
4. By the Lemma for metric space continuity, find an open ball around $x_0$ inside $f^{-1}(V)$.
5. This ball is an interval $(x_0-\delta ,x_0+\delta )$.
6. All elements of this interval are mapped into $V$, confirming the $\epsilon$-$\delta$ condition.

Assume that $f$ is continuous according to the $\epsilon$-$\delta$ definition.

1. Select any $V\subset \mathbb{R}$ open set. We must show $f^{-1}(V)$ is open.
2. Select any $x_0\in f^{-1}(V)$. Define $y=f(x_0)$.
3. Then $y_0\in V$, and so $y_0\in {\mathbf{B}}_{\epsilon }(y_0)\subset V$ for some $\epsilon >0$ by metric space continuity.
4. This ${\mathbf{B}}_{\epsilon }(y_0)$ is an interval: $y\in {\mathbf{B}}_{\epsilon }(y_0)⟺|y-y_0|<\epsilon$
5. By the $\epsilon$-$\delta$ definition of continuity, produce some $\delta >0$ so that: $|x-x_0|<\delta ⟹|f(x)-y_0|<\epsilon$
6. Therefore, whenever $x\in {\mathbf{B}}_{\delta }(x_0)$, we know $f(x)\in V$.
7. So ${\mathbf{B}}_{\delta }(x_0)$ is open (by definition), is contained in $f^{-1}(V)$, and contains $x_0$, confirming the topological continuity condition.