The Topology of Euclidean Space — Examples

Assume $\mathrm{AC}_\omega$ and $\mathrm{DC}$, let $n\ge1$, $c\in\mathbb R^n$, and $r>0$. The closed ball $\overline B_2(c,r)$ and sphere $S_2(c,r)$ are compact (For $n\ge1$, every Euclidean closed ball and every Euclidean sphere of positive radius is compact). Therefore each is closed and bounded, pseudocompact, countably compact, sequentially compact, limit point compact, and complete and totally bounded; every continuous real-valued function on either set attains both extrema (Assuming $\mathrm{AC}_\omega$ and $\mathrm{DC}$, compactness, sequential compactness, countable compactness, limit point compactness, completeness and total boundedness, pseudocompactness, closedness and boundedness, and the extreme-value property are equivalent for nonempty subsets of $\mathbb{R}^n$ with $n\ge1$). The two sets are those of Euclidean spheres and closed balls as subspaces of $\mathbb{R}^n$.

Refuted claim: every bounded subset of $\mathbb R^n$ is compact.

For $n\ge1$, the open unit ball $B_2(0,1)$ is bounded but not compact.

Refuted claim: every closed subset of $\mathbb R^n$ is compact.

For $n\ge1$, $\mathbb R^n$ is closed and unbounded, hence it is not compact.

Let $x,y\in B_2(c,r)\subseteq\mathbb R^n$, with $0\le t\le1$. The triangle inequality and absolute homogeneity give

$\lVert(1-t)x+ty-c\rVert_2\le(1-t)\lVert x-c\rVert_2+t\lVert y-c\rVert_2<r.$

So the segment from $x$ to $y$ stays in the ball. By A finite concatenation of straight segments in $\mathbb{R}^n$ is a continuous path it is a polygonal path, and $B_2(c,r)$ is polygonally connected in the sense of Polygonal paths and polygonally connected subsets of $\mathbb{R}^n$. The norm and ball conventions are those of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and Open ball, closed ball and sphere in a metric space.

The invertible $1\times1$ real matrices are the nonzero real numbers, so $GL_1(\mathbb R)=\mathbb R\setminus\{0\}$. This set is disconnected: it contains $-1$ and $1$ but not the intermediate point $0$, so it is not order-convex and cannot be connected by The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$". In contrast, For $n\ge2$, the punctured space $\mathbb{R}^n\setminus\{0\}$ is polygonally connected gives polygonal connectedness of $\mathbb R^2\setminus\{0\}$. Connectedness is understood through Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets.

False claim: every pseudocompact topological space is compact.

Let $X$ be an infinite set with a distinguished point $p$, carrying the particular-point topology. Then $X$ is pseudocompact and not compact.