# Homeomorphisms on $\mathbb{R}$

Consider the following topologies on $$\mathbb{R}$$:

$$\tau_1 = \{\mathbb{R}, \emptyset, (-n, n) \forall n\in \mathbb{Z}^+\}$$

$$\tau_2 = \{\mathbb{R}, \emptyset, [-n, n] \forall n\in \mathbb{Z}^+\}$$

$$\tau_3 = \{\mathbb{R}, \emptyset, (-r, r) \forall r\in \mathbb{R}^+\}$$

$$\tau_4 = \{\mathbb{R}, \emptyset, (-r, r)$$ and $$[-r, r] \forall r\in \mathbb{R}^+\}$$

Problem 1: Show $$\tau_1 \not\cong \tau_3$$

Problem 2: Is $$\tau_1 \cong \tau_2$$?

Problem 3: Is $$\tau_3 \cong \tau_4$$?

Note for beginners: Do not assume facts about $$\mathbb{R}$$ based on the standard Euclidean topology. These are different topologies, and so they may behave differently on otherwise familiar sets-- for example, $$(a,b)$$ and $$[a,b]$$ are non-homeomorphic interval types in the Euclidean topology, but you can't use that here. Work from the definition of Homeomorphism.

Citation: S. Morris, "Topology without Tears", problem 4.3.7 i, ii, iii

• By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions. – Lee Mosher Nov 26 '18 at 16:54
• ok, I put that in my note for beginners. – Cassius12 Nov 27 '18 at 2:32
• I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question. – Lee Mosher Nov 27 '18 at 13:12

$$|\tau_1| \neq |\tau_3|$$ while a homeomorphism induces a bijection of topologies.

Let $$f_1$$ be a bijection between $$(-1,1)$$ and $$[-1,1]$$ (standard set theory).

We can extend this to a bijection $$f_2$$ between $$(-2,2)$$ and $$[-2,2]$$ (both difference sets are also equinumerous).

So continuing this way, we get a sequence of extensions that are bijections between $$(-n,n)$$ and $$[-n,n]$$. The union of these functions is then a bijection of the reals that for each $$n$$ obeys $$f[(-n,n)] = [-n,n]$$ (and so also $$f^{-1}[-n,n]] = (-n,n)$$) and so $$f$$ is both open and continuous. So $$\tau_1 \simeq \tau_2$$.

In $$\tau_4$$ the open sets have the property:

$$\exists U,V \in \tau_4: U \subsetneq V \land |V \setminus U| =2\tag{1}$$

While $$\tau_3$$ does not have the above property as it obeys the property

$$\forall U,V \in \tau_3: U \subsetneq V \implies |V \setminus U| \text{ is infinite}\tag{2}$$

So $$\tau_3 \not \simeq \tau_4$$:

If $$f: (\mathbb{R}, \tau_4) \to (\mathbb{R}, \tau_3)$$ would be a homeomorphism, then for $$U,V$$ as in $$(1)$$ we'd have $$f[U] \subsetneq f[V]$$, both would be open and $$f[V \setminus U] = f[V] \setminus f[U]$$ would contradict $$(2)$$.