Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the following problem:

Which of the following statements are true about the open interval $(0,1)$ and the closed interval $[0,1]$?

I. There is a continuous function from $(0,1)$ onto $[0,1]$.

II. There is a continuous function from $[0,1]$ onto $(0,1)$.

III. There is a continuous one-to-one function from $(0,1)$ onto $[0,1]$.

Instead of giving a quick answer to the question, I am thinking about what is the underlying topology for the space $X_1=(0,1)$, $X_2=[0,1]$ in each of the statements.

Here are my questions:

  • What's the topology supposed to be in each of the statements? Is it the subspace topology?

  • Is it possible that all the statements are true with a suitable topology instead of using the subspace topology?

share|improve this question
1  
Similar. math.stackexchange.com/questions/42308/… –  user17762 Jun 1 '11 at 17:35

4 Answers 4

up vote 5 down vote accepted

(a) yes, the subspace topology; a.k.a. the usual topology. Since no other topology is mentioned, that is assumed.

(b) of course: any two sets with the same cardinal can be given the discrete topologies, then all three are true.

share|improve this answer
    
Fair enough. On the other hand, construction of the bijection between $[0,1]$ and $(0,1)$ is not a trivial thing, I think. The existence can be guaranteed by the Cantor–Bernstein–Schroeder theorem. –  Jack Jun 1 '11 at 22:18
2  
@Jack: There are very explicit bijections between the two based on the Hilbert hotel principle. –  Grumpy Parsnip Aug 11 '11 at 21:19

Yes, they mean the subspace topology.

Yes, it's possible that all of the statements are true with a suitable topology. Equip each of them with the trivial topology. As Wikipedia notes, two spaces carrying the trivial topology are homeomorphic (i.e. there exits a continuous bijective map whose inverse is also continuous) iff they have the same cardinality.

share|improve this answer

To add to the comments made so far:

Proposition. Let $X$ and $Y$ be nonempty sets.

  1. If $\mathscr{I}=\{\emptyset,Y\}$ is the indiscrete topology on $Y$, then for every topology $\tau$ on $X$, every map $f\colon (X,\tau)\to (Y,\mathscr{I})$ is continuous.

  2. If $\mathscr{D}=\mathcal{P}(X)$ is the discrete topology on $X$, then for every topology $\sigma$ on $Y$, every map $g\colon (X,\mathscr{D})\to(Y,\sigma)$ is continuous.

Proof.

  1. Let $f\colon(X,\tau)\to(Y,\mathscr{I})$ be any map. Then $f^{-1}(\emptyset)=\emptyset\in\tau$, $f^{-1}(Y)=X\in\tau$, so the inverse image of every open set in $(Y,\mathscr{I})$ is open in $(X,\tau)$, so $f$ is continuous.

  2. Let $g\colon(X,\mathscr{D})\to(Y,\sigma)$ be any map. Then for every $\mathscr{O}\in\sigma$, $g^{-1}(\mathscr{O})\subseteq X$, hence $g^{-1}(\mathscr{O})\in\mathscr{D}$. Thus, the inverse image of every open set in $(Y,\sigma)$ is open in $(X,\mathscr{D})$, so $g$ is continuous. QED

This is a consequence of the fact that endowing a set with the indiscrete topology is the right adjoint of the underlying set functor from $\mathcal{T}op$ to $\mathcal{S}et$, while endowing it with the discrete topology is the left adjoint.

That is: define $\mathscr{U}$ to be the functor from the category of topological spaces to the category of sets to be the forgetful/underlying set functor: given any topological space $(X,\tau)$, $\mathscr{U}(X,\tau)$ is just the set $X$; and given any continuous map $f\colon (X,\tau)\to(Y,\sigma)$ between topological spaces, $\mathscr{U}(f)$ is the set-theoretic function between $X$ and $Y$ that corresponds to $f$. This is a functor in the sense of category theory.

Define $\mathscr{I}$ to be the functor from the category of sets to the category of topological spaces defined by taking a set $X$ to the topological space $\mathscr{I}(X) = (X,\{\emptyset,X\})$, that is, $X$ endowed with the indiscrete topology; and that takes any set-theoretic map $f\colon X\to Y$ to the continuous map $\mathscr{I}(f)\colon (X,\{\emptyset, X\})\to(Y,\{\emptyset, Y\})$ that is defined on $X$ by $f$ itself. This is continuous by the first clause of the proposition above. Again, it is easy to verify that this is a functor. It is called the indiscrete topology functor.

Given a set $Y$ and a topological space $(X,\tau)$, there is a natural bijection between the set of all set-theoretic functions from $X=\mathscr{U}(X,\tau)$ to $Y$, $\mathcal{S}et(X,Y)$, and the set of all continuous functions between the topological spaces $(X,\tau)$ and $(Y,\{\emptyset,Y\})=\mathscr{I}(Y)$. That is, we have a natural bijection $$\mathcal{S}et\Bigl(\mathscr{U}(X,\tau), Y\Bigr) \stackrel{\cong}{\longleftrightarrow} \mathcal{T}op\Bigl( (X,\tau), \mathscr{I}(Y)\Bigr).$$ This says that $\mathscr{I}$ is the right adjoint of $\mathscr{U}$.

Likewise, given any set $X$, define $\mathscr{D}$ to be the function that takes the set $X$ to the topological space $\mathscr{D}(X,\mathcal{P}(X))$, that is, $X$ endowed with the discrete topology; given a set-theoretic map $f\colon X\to Y$, we get a continuous function $\mathscr{D}(f)\colon \mathscr{D}(X)\to\mathscr{D}(Y)$; this is again a functor from the category of all sets to the category of topological spaces. It is called the discrete topology functor.

Given any set $X$ and any topological space $(Y,\sigma)$, there is a natural bijection between the set of all set-theoretic functions from $X$ to $\mathscr{U}(Y,\sigma)=Y$, and the set of all continuous functions between the topological spaces $\mathscr{D}(X)$ and $(Y,\sigma)$. That is, we have a bijection $$\mathcal{T}op\Bigl( \mathscr{D}(X), (Y,\sigma)\Bigr) \stackrel{\cong}{\longleftrightarrow} \mathcal{S}et\Bigl( X, \mathscr{U}(Y,\sigma)\Bigr),$$ so that $\mathscr{D}$ is the left adjoint of the underlying set functor.

Because $\mathscr{I}$ is the right adjoint of the underlying set functor, any function into a space with the indiscrete topology is necessarily continuous. Because $\mathscr{D}$ is the left adjoint of the underlying set functor, any function from a space with the discrete topology is necessarily continuous.

Related to your question, if you endow $[0,1]$ with the discrete topology, then you can put any topology on $(0,1)$ and then use any functions whatsoever to answer the question; if you endow $(0,1)$ with the indiscrete topology, then you can put any topology on $[0,1]$ and any function $(0,1)\to[0,1]$ will be continuous.

share|improve this answer

It's worth pointing out the difference in Jesse and GEdgar answers of the second question.

The trivial topology is the least fine topologies of all where only the empty set and the whole set are open, whereas the discrete topologie is the finest of all where every subset of the space is open.

If you use one of these topologies on both intervals, then any bijection will be a homeomorphism.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.