Continuity, homeomorphism, separation axioms

In one sentence

Continuity is preimages of open sets being open, a homeomorphism is a continuous bijection with continuous inverse, and the separation axioms measure how far a space is from being metric.

Why the exam asks it

Because "continuous bijection" and "homeomorphism" are different, and because Hausdorffness is the hypothesis quietly required by half the theorems people remember from metric spaces — where it always held.

The idea

$f : X \to Y$ is continuous if $f^{-1}(V)$ is open for every open $V$. Equivalently preimages of closed sets are closed, or $f(\overline A) \subseteq \overline{f(A)}$.

Note the direction. Images of open sets need not be open — that is an open map, a separate condition. $f(x) = x^2$ maps the open $(-1,1)$ onto $[0,1)$.

A homeomorphism is a continuous bijection whose inverse is continuous. The third condition is not automatic:

• $[0,1) \to S^1$, $t \mapsto e^{2\pi i t}$, is a continuous bijection and not a homeomorphism — the inverse tears the circle at $1$.
• The identity from the discrete topology on $X$ to any coarser topology is a continuous bijection and not a homeomorphism.

It is automatic when $X$ is compact and $Y$ is Hausdorff, which is the standard rescue and worth memorising as a package.

Topological properties are those preserved by homeomorphism: compactness, connectedness, separation axioms, countability axioms. Not topological: boundedness, completeness, being a Cauchy sequence — all metric notions, which is why $(0,1)\cong\mathbb{R}$ despite one being bounded and complete-able and the other not.

The separation axioms

• $T_{0}$: distinct points are topologically distinguishable.
• $T_{1}$: each of two points has a neighbourhood missing the other $\iff$ singletons are closed.
• $T_{2}$ Hausdorff: distinct points have disjoint neighbourhoods.
• $T_{3}$ regular: a point and a closed set not containing it have disjoint neighbourhoods (plus $T_{1})$.
• $T_{4}$ normal: two disjoint closed sets have disjoint neighbourhoods (plus $T_{1})$.

Each implies the previous. Every metric space is $T_{4}$, so all of this is invisible until you leave metric spaces.

What Hausdorff buys, and each of these fails without it:

• Limits of sequences are unique.
• Compact subsets are closed.
• The diagonal is closed in $X\times X$.
• Continuous maps agreeing on a dense set are equal.

The standard non-Hausdorff space is the cofinite topology on an infinite set: closed sets are the finite ones. It is $T_{1}$ — singletons are closed — and any two non-empty open sets have cofinite complements, so they must meet. Every subset is compact, and compact subsets are emphatically not closed.

Urysohn and metrisation

Urysohn's lemma. In a normal space, disjoint closed sets $A, B$ admit a continuous $f : X \to [0,1]$ with $f|_A = 0$ and $f|_B = 1$. Tietze extends continuous functions from closed subspaces.

Urysohn metrisation theorem. A regular, second countable space is metrisable. That gives a usable test, and it is why the Sorgenfrey line — regular and not second countable — can fail to be metrisable while looking well behaved.

Normality is the axiom that misbehaves: it is not preserved by products or by subspaces. The Sorgenfrey line is normal and the Sorgenfrey plane is not.

Where intuition breaks

"A continuous bijection is a homeomorphism." $[0,1)\to S^1$. The inverse's continuity is a third condition — unless the domain is compact and the codomain Hausdorff.

"Continuous maps send open sets to open sets." Preimages, not images. $x\mapsto x^2$ on $(-1,1)$.

"Compact subsets are closed." Needs Hausdorff. In the cofinite topology every subset is compact and only the finite ones are closed.

"Limits are unique." Needs Hausdorff. In the indiscrete topology every sequence converges to every point.

"$T_{1}$ implies Hausdorff." Cofinite on an infinite set is $T_{1}$ and not $T_{2}$. The gap between the two is where most counterexamples live.

"A subspace of a normal space is normal." False; the Tychonoff plank is the classical witness. Regularity and Hausdorffness are hereditary, normality is not, and that asymmetry is examined.

"Normal spaces are closed under products." Sorgenfrey line squared. This is the cleanest counterexample and worth naming.

"Metrisable means a metric is given." It means one exists inducing the topology. Urysohn gives a sufficient condition: regular plus second countable.

The exam's angle

1. Check the direction of every continuity claim — preimage or image, open or closed.
2. For "is this a homeomorphism", test the inverse, or check compact-to-Hausdorff and get it free.
3. When a metric-space theorem is quoted, ask what replaces completeness or boundedness. Usually Hausdorff or compactness, and usually the option has dropped it.
4. Reach for cofinite first for any $T_{1}-$but-not$-T_{2}$ question, and for anything about compact sets failing to be closed.
5. For metrisability, use Urysohn — regular plus second countable — and look for a failure of second countability when the answer is no.

The night before

• Continuous $=$ preimages of open sets are open. Images need not be open.
• Homeomorphism $=$ continuous bijection with continuous inverse. $[0,1)\to S^1$ is not one.
• Compact domain $+$ Hausdorff codomain $\Rightarrow$ a continuous bijection is a homeomorphism.
• Topological: compact, connected, separation, countability. Not: bounded, complete, Cauchy. $- T_{1}$ $\iff$ singletons closed. Hausdorff gives unique limits, compact $\Rightarrow$ closed, closed diagonal.
• Cofinite on an infinite set: $T_{1}$, not $T_{2}$, every subset compact.
• Every metric space is $T_{4}$. Urysohn metrisation: regular $+$ second countable $\Rightarrow$ metrisable.
• Normality is not hereditary and not productive; Sorgenfrey line normal, Sorgenfrey plane not.