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Let $\mathcal{B} := \{\ ]a, b[ \ \times \ ]a, b[ \ | \ a, b, \in \mathbb{R}, a < b\}$ be a topological basis (I proved earlier that it is a basis), and define $\mathcal{T} = \mathcal{T}(\mathcal{B})$.

Proof that any continuous function $f: (\mathbb{R}^2, \mathcal{T}) \to (\mathbb{R}, \mathcal{T}_{eucl})$ must be constant.

I already made some progress:

As $B \in \mathcal{B}$ is open in $\mathcal{T}_{eucl}$, the identity from $(\mathbb{R}, \mathcal{T}_{eucl})$ to $(\mathbb{R}^2, \mathcal{T})$ is continious, and because $\mathbb{R}$ is connected in $\mathcal{T}_{eucl}$, it's also connected in $\mathcal{T}$.

Suppose $f$ is not constant, then there exist $a, b$ such that $f(a) > f(b)$. Now I want to find disjoint $A, B$ such that $f(a) \in A$, $f(b) \in B$ such that $f^{-1}(A), f^{-1}(B)$ are clopen, because that implies that $f(\mathbb{R}) \subset A, B$ (because $f^{-1}(A), f^{-1}(B)$ are non-empty and $(\mathbb{R}, \mathcal{T})$ is connected), which is a contradiction.

I'm just stuck finding suitable $A, B$. An open basis element will be a square that has two opposite corners on either $y = x$ or $y = -x$, but I don't really know what a closed set will look, except $\mathbb{R}^2$ minus a not-neccessarily countable union of such squares.

Here's a proof based on Rob Arthan's answer.

Assume $f$ is continuous. Notice that for any set $f^{-1}[A]$ open in $(\mathbb{R}^2, \mathcal{T})$ that contains the point $y := (-x, x)$, there is an element from the basis such that $(-x, x) \in \ ]a, b[ \ \times \ ]a, b[ \ $, so $a < -x$ and $b > x$, so $[-x, x] \times [-x, x] \subset \ ]a, b[ \ \times \ ]a, b[ \ \subset f^{-1}(A)$.

Let $p \in (\mathbb{R}^2, \mathcal{T})$ be arbitrary. We can choose $x$ large enough such that $p \in f^{-1}[A]$ as defined previously. Now take an open $A$ which contains $f(y)$. Then $f^{-1}[A]$ will be open as $f$ is continuous and contain $y$, justifying calling that set $f^{-1}[A]$. Then $p, y, (0, 0) \in f^{-1}[A]$. So $f(p), f(y), f((0, 0)) \in A$. This is true for all open $A$ containing $f(y)$, so in particular $]f(y) - \frac{1}{n}, f(y) + \frac{1}{n}[$ and the only number contained in this for all $n$ is $f(y)$, so $f(p) = f(x) = f((0, 0))$. So $f(p) = f((0, 0))$ for all $p$, so $f$ is constant.

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    $\begingroup$ I'm not seeing how $\mathcal{T}$ is different from the usual topology of $\mathbb{R}^2$, which does not have this property... $\endgroup$ – Ian Jan 28 '18 at 16:18
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    $\begingroup$ @Ian, for example $A := ]0, 1[ \times ]2, 7[$ is open in the usual topology, but not in $\mathcal{T}$ as there is no basis element contained in $A$. (Suppose there is, then $]a, b[ \subset ]0, 1[$, but also $]a, b[ \subset ]2, 7[$. $\endgroup$ – Pel de Pinda Jan 28 '18 at 16:23
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    $\begingroup$ Oh, yes, $(a,b) \times (a,b)$ as opposed to $(a,b) \times (c,d)$, got it. $\endgroup$ – Ian Jan 28 '18 at 16:26
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Hint: an alternative approach is to use the fact that your space $\cal B$ is very far from Hausdorff. If $x > 0$, then any open set containing the point $(-x, x)$ contains the entire closed square $S_x = [-x, x] \times [-x , x]$. Now given any $p \in \Bbb{R}^2$, for large enough $x$, $p \in S_x$, which implies $f(p) = f(x) = f(0)$ (since if $A$ is any open set containing $f(x)$, $f^{-1}(A) \supseteq S_x \supset \{p, x, 0\}$, so $\{f(p), f(x), f(0)\} \subseteq A$).

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  • $\begingroup$ I think I solved it, would you mind checking it? It's in the main post. Thanks for your help. $\endgroup$ – Pel de Pinda Jan 28 '18 at 18:46
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    $\begingroup$ My pleasure! Your write-up looks good to me. $\endgroup$ – Rob Arthan Jan 28 '18 at 19:04

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