Let $E$ be the open interval $(0, 1)$ with the usual metric topology and $\Bbb R$ be given the usual topology.

Let $\Bbb R^J$ be given the topology of pointwise convergence $T_p$, where the index set $J$ comprises all continuous functions $f : E \to \Bbb R$.

Define a function $T : E \to \Bbb R^J$ by $T(x) = (f(x))_{f∈J}$ . Suppose that the image set $T(E)$ is given the subspace topology as a subspace of $(\Bbb R^J, T_p)$.

I want to show that $T$ is a homeomorphism from $E$ onto $T(E)$ but am not sure what method to employ even hmm.

Is there an explicit map to find? or find continuous maps $f$ and $g$ s.t. $f\circ g=g\circ f= i$ where $i$ is the identity map?


First notice that $T : E \to T(E)$ is continuous. Indeed, let $(x_\lambda)_{\lambda\in\Lambda}$ be a net in $E$ converging to $x \in E$. Then also $f(x_\lambda) \to f(x)$ for every $f \in J$ so $$T(x_\lambda) = (f(x_\lambda))_{f \in J} \to (f(x))_{f \in J} = T(x)$$ in the product topology on $\mathbb{R}^J$.

We can construct the inverse for $T$. The inverse $T^{-1} : T(E) \to E$ is given by the canonical projection $\pi_g : \mathbb{R}^J \to \mathbb{R}$ (with appropriate domain and codomain) where $g \in J$ is the inclusion $g(x) = x, \forall x \in E$.

Indeed, for any $x \in E$ we have $$T^{-1}(T(x)) = T^{-1}((f(x))_{f \in J}) = \pi_g(f(x))_{f \in J} = g(x) = x$$ and for any $T(x) \in T(E)$ with $x \in E$ we then have $$T(T^{-1}(T(x))) = Tx$$ which shows that $T^{-1}$ is indeed the inverse for $T$. It is clearly continuous so we conclude that $T$ is a homeomorphism.

  • $\begingroup$ Ah thank you! By constructing the inverse for T like that, does it suffice to prove T is an open map though? $\endgroup$ – Homaniac Mar 7 at 0:19
  • $\begingroup$ @Homaniac $T$ is a homeomorphism if both $T$ and $T^{-1}$ are continuous. Equivalently, $T$ must be continuous and an open map. $\endgroup$ – mechanodroid Mar 7 at 0:21
  • $\begingroup$ Ah thanks makes sense, I am needing further details/comments too for a suggested solution in another homeomorphism question,if you like to help :) math.stackexchange.com/questions/3129188/… $\endgroup$ – Homaniac Mar 7 at 0:28
  • $\begingroup$ @Homaniac The existing answer already provides a pretty explicit homeomorphism. Do you have problems with showing it is a homeomorphism, or would you like a different map (not that one comes to mind now, I'll have another look tomorrow)? $\endgroup$ – mechanodroid Mar 7 at 0:33
  • $\begingroup$ More of the showing that is a homeomorphism and how it matches up with the equivalence relation conditions given in that qns though I guess this only goes under the comments section over there, not really a full answer but thanks for the help~ $\endgroup$ – Homaniac Mar 7 at 0:38

You have the map $T$. This is the homeomorphism between $E$ and and $T[E]$.

So you need to show that $T$ is 1-1/injective. This comes down to the fact that if $x \neq y$ in $E$ there is a continuous function $f: E \to \mathbb{R}$ (so $f \in J$) such that $f(x) \neq f(y)$.

Then $T$ is shown to be a bijection between $E$ and $T[E]$.

$T$ is continuous because $\pi_f \circ T = f$ is continuous for every $f \in J$.

$T$ has a continuous inverse too. This is the trickest part. This turns out to be equivalent to the fact that if $C \subseteq E$ is closed and $x \notin C$, there is a function $f \in J$ such that $f(x) \notin \overline{f[C]}$. But in this case you might be able to find a more specific direct argument, if you look for it

But this argument in fact works not just for $E$ but for any completely regular $T_1$ space: these all embed into a power of reals.

  • $\begingroup$ In the part for showing T has a continuous inverse, is the equivalence of it to that other fact a standard result? Otherwise, I am having difficulty giving a direct proof for T having a continuous inverse~ $\endgroup$ – Homaniac Mar 6 at 18:00
  • $\begingroup$ @Homaniac it’s a standard result in some books (Engelking general topology ) and in my old university notes. I’ve proved it on this site too. $\endgroup$ – Henno Brandsma Mar 6 at 18:08
  • $\begingroup$ Thanks I see, do you have any links? I'll try to look it up~ $\endgroup$ – Homaniac Mar 6 at 18:13
  • $\begingroup$ @Homaniac 34.2 in Munkres is one place. $\endgroup$ – Henno Brandsma Mar 6 at 18:40
  • $\begingroup$ @Homaniac this thread on another forum has the proof too. $\endgroup$ – Henno Brandsma Mar 6 at 18:45

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