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From Wikipedia:

"Let $K$ be a topological field, namely a field with a topology such that addition, multiplication, and division are continuous. In most applications $K$ will be either the field of complex numbers or the field of real numbers with the familiar topologies."

I have two questions:

Can you give me an example of a topology on $K$ such that addition is not continuous? I assume that addition is continuous means that $f(x,y) = x + y$ is continuous in both $f(x, \cdot)$ and $f(\cdot, y)$.

The other question is: how is division continuous? I assume again that division by $y$ is a function $f_y(x) = \frac{x}{y}$. How is $f_0$ continuous in the standard topology?

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Continuous where defined. And it is more than separately continuous, it is continuous as a function of two variables, product topology. – André Nicolas Apr 28 '12 at 6:47
up vote 3 down vote accepted

Sure, you can put a terrible topology on $K$ and then addition won't be continuous. For example, pick the cofinite topology on $\mathbb{R}$ if you like. The preimage of the origin, which is a closed point, under addition $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ will be the antidiagonal set $\{(x, -x)\} \subset \mathbb{R}^2$, which is infinite and therefore not closed.

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To clarify, continuity of addition means that the addition map $X \times X \to X$, where $X \times X$ gets the product topology, is continuous – user29743 Apr 28 '12 at 6:52
This example shows that it's not enough just to consider the functions $f(x, .)$ as $x$ varies, since those will be continuous for this topology. – user29743 Apr 28 '12 at 6:53
$f(x,\cdot)$ and $f(\cdot,y)$ will be continuous for every topology on $X$ since the function with one argument fixed is a translation, right? Or are there topologies that make a translation map discontinuous? – Rudy the Reindeer Apr 28 '12 at 18:56
@Clark Kent there are definitely topologies making translations discontinuous. Basically, pick your favorite set-theoretic bijection of $\mathbb{R}$ with itself - maybe fix the natural numbers and slide the open (n, n+1) intervals to the right. Now topologize $\mathbb{R}$ via this bijection (so a set is open iff its image under the bijection is open). This is a disgusting topology with respect to which no reasonable map can be continuous. – user29743 Apr 28 '12 at 19:00

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