# Is $f(x) = (x,x)$ not continuous in this topology?

Let $$S$$ be the unit circle with the usual topology and let $$\mathbb{R}$$ be equipped with the discrete topology, then consider $$S \times \mathbb{R}$$ with the product topology, and consider the map $$f(x) = (x,x)$$ from $$S$$ to $$S \times \mathbb{R}$$, where here we are identifying $$S$$ with, say, $$[0,1)$$.

Is this map $$f$$ continuous?

The preimage of any $$U \times R$$ under $$f$$ for $$U$$ open in $$S$$ and $$R$$ open in $$\mathbb{R}$$ (any subset) is just $$U$$ or $$\emptyset$$, which is open in $$S$$, so it seems that $$f$$ should be continuous.

On the other hand...

Consider a non-closed subset $$B$$ of $$S$$. We have that $$B = f^{-1}(f(B))$$, but $$f(B)$$ is closed since any converging sequence in $$S \times \mathbb{R}$$ must be eventually constant. So, if $$f$$ were continuous we would have $$B$$ is closed since it is the inverse image of a closed set under a continuous map, a contradiction.

Clearly there is a mistake somewhere, but where have I gone wrong? Is the map $$f$$ continuous or not?

• The preimage of $U \times R$ is actually $U \cap R$. Feb 24, 2021 at 0:22
• Alternately, $f$ is continuous if and only if $\pi_1 \circ f$ and $\pi_2 \circ f$ are continuous where $\pi_1 : S \times \mathbb{R} \to S$ and $\pi_2 : S \times \mathbb{R} \to \mathbb{R}$ are the projections. $pi_1 \circ f$ is the identity function $S \to S$ so there's no problem there. The issue is: is $\pi_2 \circ f : S \to \mathbb{R}$ continuous? Feb 24, 2021 at 0:24
• Also, converging sequences in $S \times \mathbb{R}$ don't necessarily have to be eventually constant. It's sufficient (and also necessary) for the first component to be a convergent sequence in $S$, and the second component to be a convergent sequence in $\mathbb{R}$ and therefore eventually constant. Feb 24, 2021 at 0:26
• $S\times \{0\}$ is open in $S\times \mathbb R$, but $f^{-1}(S\times\{0\})$ is a single point. Feb 24, 2021 at 0:47

To see that $$f$$ is not continuous, observe that $$f[S]$$ is a discrete subset of $$S\times\Bbb R$$. To see this, let $$x\in S$$, where I identify $$S$$ with $$[0,1)$$; then $$f(x)=\langle x,x\rangle$$, so $$S\times\{x\}$$ is an open nbhd of $$f(x)$$ that contains no other point of $$f[S]$$. (In fact the discrete subsets of $$S\times\Bbb R$$ are precisely those $$A\subseteq S\times\Bbb R$$ with the property that $$A\cap(S\times\{x\})$$ is finite for each $$x\in\Bbb R$$.) Since $$S$$ is not a discrete space, $$f$$ cannot be continuous.