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Having infinitely many open sets in a topology is a topological property
#1: What are example of two topological spaces, preferably simple ones, which this statement show are not homeomorphic.
But, more importantly why does it show this? Having trouble understanding topological properties in connection to homeomorphisms.
#2: What about two spaces where this statement doesn't help deciding whether they are homeomorphic or not?
If $X$ and $Y$ are topological spaces with respective topologies $\tau_X$ and $\tau_Y$, and if $f : X \to Y$ is a homeomorphism, then $f$ induces a bijection $\tau_X \mapsto \tau_Y$: each $U \in \tau_X$ is mapped to its image $f(U) \in \tau_Y$ defined as usual by the formula $$f(U)=\{f(x) \,|\, x \in U\} $$ The proof that this map $\tau_X \to \tau_Y$ is a bijection is simple, and uses only the definition of homeomorphism.
As a consequence, if $X$ and $Y$ are homeomorphic then the two sets $\tau_X$ and $\tau_Y$ have equal cardinalities. So, for example, it cannot happen that one is finite and the other is infinite. It also cannot happen that one is countably infinite and the other is uncountably infinite.
Space 1: the interval $[0, 1]$ with the usual topology. Space 2: the interval $[0, 1]$ with the indiscrete topology, in which the only open sets are the empty set and the whole set.
Note that as sets, these spaces are identical; as topological spaces, they're incredibly distinct
Why is a set with a finite topology different from a set with an infinite topology? A homeomorphism $f$ has to map open sets to open sets (and its inverse, $g$ must do so in the other direction, with $f \circ g = id_X$ and $g \circ f = id_Y$). If $X$ has an infinite topology and $Y$ has a finite one, there must be an open set $U$ in $Y$ that's the image of more than one open set in $X$, say $V_1$ and $V_2$. But what's the preimage $g(U)$ of $U$ in $X$? Under the homeomorphism, it must be both $V_1$ and $V_2$, and that's impossible, because these were distinct open sets of $X$?
Take $\mathbb Z$ with the usual topology and with the topology$$\tau=\{\emptyset,\mathbb{Z}\}\cup\bigl\{\{a,a+1,a+2,\ldots\}\,|\,a\in\mathbb Z\bigr\}.$$This topology is countable, but the usual one isn't (since every subset of $\mathbb Z$ is open).
Note that if $f$ is a homeomorphism from a topological space $(X_1,\tau_1)$ onto a topological space $(X_2,\tau_2)$, then $f$ induces a bijection between $\tau_1$ and $\tau_2$:$$\begin{array}{ccc}\tau_1&\longrightarrow&\tau_2\\X&\mapsto&f(X).\end{array}$$Therefore, $\#\tau_1=\#\tau_2$.
Let $X$ be a point and $Y$ the integers, both with the discrete topology. Then $X$ has finitely many open sets, while $Y$ has infinitely many open sets. But since there's no bijection between the two, they definitely can't be homeomorphic.
Topological properties are the same as invariants under homeomorphism. Checking for different topological properties can tell you when spaces are not homeomorphic.
