If we look at any set $X$ with the trivial topology, then all functions from $X$ to $X$ are continuous. We could also take the discrete topology and get the same result: all functions are continuous.

Another example: Take the Sierpinski space, all functions from it to itself, except the function that switches 0 and 1, are continuous. Again there is another topology that gives the same continuous functions (just take the other singleton as open).

Is this true for $\mathbb{R}$ aswel? Note that I'm not asking if $\mathbb{R}$ is completely regular, which is a seemingly similar but different property (In that property, the image space is equipped with the Euclidean topology if I'm not mistaken).

I'm interested in $\mathbb{R}$ especially because this would give another way to think about continuous functions on it. High school students learn an epsilon delta definition but you can teach them what an open set is and define continuous that way. The usual definition of open is then "union of open intervals", but maybe there is a different choice of opens to get the same notion of continuous functions.

That being said, I'm also interested in other spaces on which the topology can be changed to obtain the same continuous functions to itself, like the examples I gave, or in spaces where you can show that the topology is the only one that gives those continuous maps.

Thanks in advance


Assume $X$ has more than 1 element. The argument that CarryonSmiling makes can be generalised as follows:

Let's call a space that is T1 and connected a Tc space. If two topologies on $X$ have the same continuous maps from $X$ to itself, then they are either both Tc or both not Tc. This is true because Tc is equivalent to the property that CarryonSmiling uses, that is expressed only with continuous maps from $X$ to itself:

$$(*)\quad\mbox{No map $h:X\rightarrow X$ whose range has exactly two elements is continuous.}$$

To see this equivalency, note that $$T1 \iff \mbox{closed singletons} \iff \mbox{every subspace with exactly 2 elements is discrete}$$ $$\mbox{connected} \iff \mbox{every continuous map to the discrete space with 2 elements is constant}$$ That means that T1 and connectedness together are exactly $(*)$. Maybe this can help to find a space that is determined by its continuous maps to itself. Please let me know if any part of this reasoning is invalid.

  • $\begingroup$ What happens if we consider the topology with sub-basis $\tau\cup \{\mathbb Q \}$ $\endgroup$ – Jorge Fernández Hidalgo Aug 12 '16 at 18:57
  • $\begingroup$ @CarryonSmiling I think then the usual function discontinuous on exactly $\mathbb{Q}$ (namely: map $x$ to $0$ if $x$ is irrational, and $x$ to ${1\over q}$ if $x={p\over q}$ in lowest terms) winds up being continuous in the new topology? $\endgroup$ – Noah Schweber Aug 12 '16 at 18:59
  • $\begingroup$ nvm, $f(x)=x+\sqrt{2}$ is not continuous. $\endgroup$ – Jorge Fernández Hidalgo Aug 12 '16 at 19:03
  • 1
    $\begingroup$ If you make $\mathbb{Q}$ closed, you will have to make every dense countable subset closed as well, just to make sure all homeomorphisms still work. But the intersection of two dense countable subsets is not necessarily dense, so presumably all countable subsets would have to be closed. $\endgroup$ – Niels J. Diepeveen Aug 14 '16 at 13:48
  • 2
    $\begingroup$ But that won't work either. There is a continuous function $f$ such that $f^{-1}[1/n] = [(2n+1)^{-1}, (2n)^{-1}]$ for all $n \in \mathbb{N}$. For such a function the inverse image of the countable set $\{ 1/n \mid n \in \mathbb{N} \}$ is not the union of a closed set and a countable set. $\endgroup$ – Niels J. Diepeveen Aug 15 '16 at 12:22

Carry on Smiling has shown that if $\tau$ is a topology on $\mathbb{R}$ has the same continuous functions $\mathbb{R}\to\mathbb{R}$ as the usual topology, then $\tau$ contains the usual topology. I will now show that $\tau$ must also be contained in the usual topology, and thus must actually be equal to the usual topology.

Suppose $A\subset\mathbb{R}$ is a set that is $\tau$-closed but not closed. Choose a point $x\in\mathbb{R}\setminus A$ and a sequence of points $(x_n)$ in $A$ which converge to $x$. To simplify notation, let me assume $x=0$ and $x_n=1/n$. In fact, this loses no generality: we can assume the $x_n$ are converging to $x$ monotonically by passing to a subsequence, and then we can apply a homeomorphism of $\mathbb{R}$ to send $x$ to $0$ and $x_n$ to $1/n$.

So we may assume $0\not\in A$ but $1/n\in A$ for all positive integers $n$. Now consider the map $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=0$ for $x\leq 0$, $f(x)=1/n$ if $x\in [1/(2n+1),1/2n]$ for some positive integer $n$, $f(x)=1$ if $x\geq 1/2$, and interpolate linearly for other values of $x$. Then $f$ is continuous. Let $g$ be defined similarly to $f$, except that $g(x)=1/n$ if $x\in [1/2n,1/(2n-1)]$. Let $h(x)=f(-x)$ and $i(x)=g(-x)$. Now observe that $f^{-1}(A)\cup g^{-1}(A)$ contains all positive numbers, and $h^{-1}(A)\cup i^{-1}(A)$ contains all negative numbers. On the other hand, none of these sets contain $0$, since all the functions map $0$ to $0$ and $0\not\in A$. Thus $f^{-1}(A)\cup g^{-1}(A)\cup h^{-1}(A)\cup i^{-1}(A)=\mathbb{R}\setminus\{0\}$.

By assumption, $f$, $g$, $h$, and $i$ are all $\tau$-continuous. It follows that $\mathbb{R}\setminus\{0\}$ is $\tau$-closed. For each $r\in\mathbb{R}$ $j_r(x)=x-r$ is also $\tau$-continuous, so in fact $\mathbb{R}\setminus\{r\}$ is $\tau$-closed for all $r$. But this means $\tau$ is the discrete topology, which is impossible since then every function would be $\tau$-continuous.

Thus no such set $A$ can exist, and $\tau$ is contained in the usual topology.

  • $\begingroup$ Is it possible that there are some mistakes in your notation? I really don't understand your definition of f. I think some n's should be x's no? $\endgroup$ – Jens Renders Aug 18 '16 at 7:41
  • $\begingroup$ Oops, should be fixed now. $\endgroup$ – Eric Wofsey Aug 18 '16 at 7:45
  • $\begingroup$ Thanks! I finally had the time to go trough it and I get it now. BTW, the definitions of h,i and j are not needed, with f and g alone you get that ]0,infinity[ is closed which makes it disconnected, but it should be connected. $\endgroup$ – Jens Renders Aug 19 '16 at 10:09

Suppose $\tau$ is such a topology. For clarity, I'll use "open" to mean open in the usual sense, and "$\tau$-open" to mean open in the new topology, $\tau$; similarly with "continuous" and "$\tau$-continuous."

We'll show that $\tau$ is in fact a refinement of the usual topology on $\mathbb{R}$. The only fact about the usual topology used is $$(*)\quad\mbox{No map $h:\mathbb{R}\rightarrow\mathbb{R}$ whose range has exactly two elements is continuous.}$$

We begin by showing that $\tau$ is $T_1$. Take $a\neq b$, and take two non-empty subsets $A,B$ with $A\cap B=\varnothing$ and $A\cup B=\mathbb R$ and with $B$ $\tau$-open. Such $A, B$ exist since $\tau$ is not indiscrete: if $\tau$ were, it would have too many continuous functions.

Now by $(*)$, the function that sends $A$ to $a$ and $B$ to $b$ is not continuous in the usual topology, so it is not $\tau$-continuous either.

Since $f$ isn't $\tau$-continuous, we must have that $A$ is not $\tau$-open. Now, the only $f$-preimages are $\emptyset, B, \mathbb{R}, A$; the only one of these which is not $\tau$-open is $A$, so for $f$ to not be $\tau$-continuous there must be some $\tau$-open set $U$ with $f^{-1}(U)=A$. That is, there must be a $\tau$-open set that contains $a$ and not $b$.

So we have that $\tau$ is $T_1$. (Note: $\tau$ is also connected by the same argument, but that's not directly useful for the rest of this argument.)

We can now prove that $\tau$ refines the usual topology. Since $\tau$ is $T_1$, we know that $\mathbb R\setminus \{x\}$ is $\tau$-open for all $x$. For $a\in\mathbb{R}$, consider the function $f_a(x)=\max(a,x)$. This function is continuous and hence $\tau$-continuous. Since $\mathbb R\setminus \{a\}$ is $\tau$-open, its $f_a$-preimage must be $\tau$-open; that is, $(a,\infty)$ is $\tau$-open. We can prove $(-\infty,a)$ is open analogously. Intersecting open sets, we get that $(a, b)$ is $\tau$-open for every $a,b\in\mathbb{R}$.

  • $\begingroup$ the function is not continuous, and the only preimages are $\varnothing,A,B,\mathbb R$. And we have that all three except $A$ are open. $\endgroup$ – Jorge Fernández Hidalgo Aug 12 '16 at 18:15
  • $\begingroup$ Oh, I see the confusion, we take $B$ so that $B$ is open in the new topology (I am implicitly assuming the new topology is not indescrete) $\endgroup$ – Jorge Fernández Hidalgo Aug 12 '16 at 18:31
  • $\begingroup$ Ah, that helps! Do you mind if I edit for clarity? (I want to replace a couple of the uses of "open" with "$\tau$-open," where $\tau$ denotes the new topology; I have some trouble reading this otherwise.) $\endgroup$ – Noah Schweber Aug 12 '16 at 18:34
  • $\begingroup$ Sure, knock yourself out. $\endgroup$ – Jorge Fernández Hidalgo Aug 12 '16 at 18:35
  • 1
    $\begingroup$ For the function to be $\tau$-discontinuous there has to be an open set $U$ with $f^{-1}(U)$ not open, but the only possibilities for $f^{-1}(U)$ are $A,B,\varnothing$ and $\mathbb R$, we know the last three are open, so $A$ cannot be $\tau-open$ $\endgroup$ – Jorge Fernández Hidalgo Aug 18 '16 at 4:00

Your Answer

