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One of my homework problems is this: "Let X = $\mathbb{R} \!\,$ with the usual metric and let X′ be a discrete metric space. Describe all continuous functions from X to X′."

A function f : X $\rightarrow$ X' is continuous if for all $\epsilon$>0, there exists a $\delta$>0 s.t. for all x,y $\in \!\,$X, if d(x,y)

The most common discrete metric we discuss is this:

d(x,y) = 0 if x=y &

d(x,y) = 1 if x$\ne$y

My two ideas for continuous functions f : X $\rightarrow$ X' are:

1) f(x) = k where k is a constant

2) f(x) = [x]

My reasoning is that both of these functions take two values x,y in $\mathbb{R} \!\,$ and return f(x) and f(y) s.t. f(x)=f(y), so the discrete metric has distance of 0. For 1, this is true for all x,y. For 2, this is true for d(x,y) < min {d(x,1/2), d(y,1/2)}.

Do both of these work? Are there any more such functions?

Thanks!

Jeff
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2 Answers2

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Hint : discrete metric is totally disconnected. but $\mathbb{R}$ is connected.

Do you know this result : Continuous image of a connected set is connected?

GA316
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  • We are not being introduced to connectedness until the next section. – Jeff Nov 24 '13 at 04:10
  • then it is right time to see the definition of connectedness. You want to prove using the definition of continuous functions? – GA316 Nov 24 '13 at 04:15
  • This exercise follows immediately after this theorem: "A function f : X → X′ is continuous iff for any open set V ⊂ X′, the set f_inverse(V) is an open set in X", this corollary: "Suppose that (X, d), (X′, d′), and (X′′, d′′) are metric spaces, and f : X → X′ and g : X′→ X′′ are continuous. Then g ◦ f : X → X′′ is continuous", and this exercise, which I have already done: "Let X and X′ be metric spaces and assume that X has the discrete metric. Show that any function f : X → X′ is continuous." I'd like to do my problem by using these 3 results and the defn. of continuity, if possible. – Jeff Nov 24 '13 at 04:20
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Since everything is open in $\;X'\;$ , the inverse image of everything in $\;X\;$ must be open.

Check now that if there exist $\;x,y\in X\;$ s.t. $\;f(x)\neq f(y)\;$, then since $\;U:=\{f(x),f(y)\}\subset X'\;$ is open, so must be $\;f^{-1}(U)\subset X\;$, and you surely know what are the open set in $\;\Bbb R\;$ with the usual topology...get a contradiction.

DonAntonio
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  • Why is "Since everything is open in X′ , the inverse image of everything in X must be open," true? The definition of an inverse image is apparently f_inverse[U] = {x belongs to X | f(x) belongs to U}. I don't see the immediate connection. – Jeff Nov 24 '13 at 05:20
  • If the function $;f:X\to X";$ is continuous, then $;f^{-1}(U);$ (yes, the definition of "inverse image" is what you wrote) must be open for any open $;U\subset X';$, and since any subset in $;X';$ is open then the inverse image of anything must be open in $;X=\Bbb R;$ with the usual Euclidean topology... – DonAntonio Nov 24 '13 at 12:23
  • Sorry I couldn't understand this proof. I will just have to see what the feedback on the homework says. – Jeff Nov 25 '13 at 08:23
  • @DonAntonio Hi, any chance you can elaborate what is the contradiction ? we have $f(a,b) = U$ , i don't see what's wrong –  Mar 20 '17 at 10:29
  • @Liad Well, this was a looooong time ago, but the arguments are simple: if $;f(x)=a\neq b=f(y);$ , and since $;U:={,a,,b,}\subset X';$ is open (as everything's open in a discrete space!) , then it must be that $;f^{-1}(U)\subset X;$ is open if $;f;$ is continuous . Yet the only open subsets in $;X=\Bbb R;$ (the real line with the usual topology) are the unions of open intervals (and then use the Intermediate Value Theorem)...Of course, connectedness can also be used as $;X';$ is totally disconnected, as the other answer mentions. – DonAntonio Mar 20 '17 at 10:45
  • @DonAntonio thanks , indeed it was a long time ago :). i understand that everything is open in $X'$ , but $f:R \to X'$ , so we have that the image of union of open intervals is $U = {f(x) , f(y) }$ for $f(x) \ne f(y)$. how can i use intermediate value theorem? the image of $f$ is not contained in $R$ (im new to topo. so im not familiar with disconnected term) –  Mar 20 '17 at 10:50
  • I'm not sure I understand you: how come the image of $;f;$ is not contained in $;\Bbb R;$ ? It is contained in $;X'\subset\Bbb R;$, with the discrete topology... And, in fact, I think that, in the last analysis, to continue with this proof I would need to use in some form connectedness, as in the other answer but in a way much messier trajectory... – DonAntonio Mar 20 '17 at 11:16
  • @DonAntonio $X'$ is a discrete metric space - who said it is contained in $R$ ? (sorry for all the questions .. im trying to understand this) –  Mar 20 '17 at 11:17
  • @DonAntonio btw, the hint i have in my exercise is to show $R$ is a union of 2 distinct open interval in the case $f$ is not constant. –  Mar 20 '17 at 11:20
  • @Liad Oh, now I see....I thought $;X';$ was $;\Bbb R;$ itself with the discrete topology. Please disregard my commentaries with this...hehe. Indeed, too long a time has ellapsed...Anyway, using your hint: suppose $;f;$ isn't constant and etc., as before. Put $;U:={x\in X;/;f(x)=a\in X'};,;;V:={x\in X;/;f(x)\neq a};$ . By assumption, $;U,V;$ are open and non-empty, and $;\Bbb R=f^{-1}(U)\cup f^{-1}(V);$ , with $;f{-1}(U),,f^{-1}(V);$ also disjoint and open . This cannot be. – DonAntonio Mar 20 '17 at 11:35
  • Let us continue this discussion in chat. –  Mar 20 '17 at 19:34