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We say two topologies $\tau$ and $\rho$ on $X$ are similar if the set of continuous functions $f:(X,\tau) \rightarrow (X,\tau)$ is the same as the set of continuous functions $f:(X,\rho)\rightarrow (X,\rho)$.

Does there exist a topology $\tau$ that is similar to the euclidean topology on $\mathbb R$?

This was asked here but all we could prove is that $\tau$ must be a refinement of the euclidean topology.

Regards.

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    $\begingroup$ The proof used the construction of the Euclidean topology from the subbasis of sets of the form $(-\infty, a)$ and $(a, \infty)$, so it was specific to the Euclidean topology, not a general fact. $\endgroup$ – user44191 Aug 18 '16 at 3:05
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    $\begingroup$ Never mind, my question clearly has the answer "no", by considering the discrete and indiscrete topologies. $\endgroup$ – user44191 Aug 18 '16 at 3:31
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    $\begingroup$ @RyanBudney I am very interested in knowing what the standard answer is. $\endgroup$ – Jorge Fernández Aug 18 '16 at 4:02
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    $\begingroup$ I am surprised by the vote to close. @RyanBudney, I am also curious about the standard answer, which I have certainly not seen before . . . $\endgroup$ – Noah Schweber Aug 18 '16 at 4:38
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    $\begingroup$ @RyanBudney I agree with the original poster and Noah. In particular I'd be curious to know if there is a comprehensible criterion for an arbitrary space. It's certainly not unique in general (e.g., the example of user44191 of the indiscrete versus discrete topology). $\endgroup$ – Gabriel C. Drummond-Cole Aug 18 '16 at 5:28
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The only topology similar to the Euclidean topology on $\mathbb{R}$ is the Euclidean topology.


Suppose there is such a topology $\tau$. I'll use "open," "continuous," etc. to mean with respect to the Euclidean topology and "$\tau$-open" etc. for $\tau$.

Since $\tau$ is a refinement of the Euclidean topology, there exists a point (without loss of generality $0$) and a $\tau$-open neighborhood $U$ which does not contain any open interval containing $0$. Then there is a sequence $x_i$ in the complement of $U$ converging to $0$. For convenience choose a monotonic subsequence $y_i$, without loss of generality positive, and suppose $y_1=1$.

Now consider the function $\mathbb{R}\to\mathbb{R}$ which fixes the complement of $(0,1)$, takes $[\frac{1}{2n},\frac{1}{2n-1}]$ to $y_n$, and takes $[\frac{1}{2n+1},\frac{1}{2n}]$ to $[y_{n+1},y_n]$ linearly. This function is continuous, hence $\tau$-continuous.

Then the preimage of $(-1,1)\cap U$ is a $\tau$-open subset $V$ of $(-1,0] \cup \bigcup_n (\frac{1}{2n+1},\frac{1}{2n})$ which contains $0$.

There is a homeomorphism (hence a $\tau$-homeomorphism) $\phi$ of $\mathbb{R}$ which fixes the complement of $(0,2)$, takes $[1,2]$ to $[\frac{1}{2},2]$ linearly, and takes $[\frac{1}{k+1},\frac{1}{k}]$ to $[\frac{1}{k+2},\frac{1}{k+1}]$ linearly.

Then $\phi(V)\cap V$ is a $\tau$-open subset of $(-1,0]$ containing $0$.

But then $(-\infty,0]$ is $\tau$-open, so that $\mathbb{R}$ is not $\tau$-connected, a contradiction since then there is a $\tau$-continuous map sending $(-\infty,0]$ to one point and $(0,\infty)$ to another.

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This is a special case of much more general results surveyed in the book MR0393330 Magill, K. D., Jr. A survey of semigroups of continuous selfmaps. Semigroup Forum 11 (1975/76), no. 3, 189–282.

For example, Theorem 2.3 of this book says that only the abstract semigroup structure (with respect to composition) of the set of continuous maps is sufficient to recover the topology.

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    $\begingroup$ How can what you say be correct, when, as mentioned, the trivial and discrete topologies give the same set of continuous functions? $\endgroup$ – Sam Hopkins Aug 18 '16 at 15:20
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    $\begingroup$ Okay, I see there are some conditions in this theorem. The spaces have to be "generated" (see Definition 2.2). $\endgroup$ – Sam Hopkins Aug 18 '16 at 15:22
  • $\begingroup$ Real line is "generated":-) $\endgroup$ – Alexandre Eremenko Aug 18 '16 at 15:31
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    $\begingroup$ $\mathbb R$ is "generated", but it is not obvious to me (without answering the above question) that $(\mathbb R,\tau)$ is "generated". $\endgroup$ – Keith Kearnes Aug 18 '16 at 18:35
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    $\begingroup$ Agreed. I'm not sure this theorem really answers the general question of when, given a topological space $(X,\tau)$ there is another topology $\tau'$ on $X$ with the same continuous functions. $\endgroup$ – Sam Hopkins Aug 19 '16 at 0:44

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