Why are there no “continuous maps” in algebra. [closed]

Or maps that behave similarly? Sorry if this is a strange question.

• Can you elaborate on the kind of behavior you're talking about? – Malice Vidrine Apr 23 '16 at 15:56
• All polynomial maps are continuous for the Zariski topology. Actually there is a lot of topology in abstract algebra : $I$-adic topology on modules, Zariski topology on algebraic varieties, Krull topology on Galois groups,etc. – Captain Lama Apr 23 '16 at 15:57
• Continuing with what Captain Lama began: the usual, inherited Euclidean topology in $\;\Bbb R^{n^2}\;$ or $\;\Bbb C^{n^2}\;$ for the set (algebra) of square matrices of degree $\;n\;$ , or the usual Euclidean topolgy for any $\;n-$ dimensional space $\;V\cong\Bbb R^n\,,\,\,or\,\,\,\Bbb C^n\;$ , etc. – DonAntonio Apr 23 '16 at 16:03
• Or one can simply talk about topological groups and, for example, the Peter-Weyl theorem. – anomaly Apr 23 '16 at 16:07
• Also, with finite algebraic structures, one can make all the algebraic operations continuous by invoking the discrete topology, but this isn't all that interesting since all the sets are open (continuity isn't a very useful concept in this context). – David Wheeler Apr 23 '16 at 16:32

It is not correct to say that there are no continuous maps in algebra. For instance, if you take a finite dimensional real vector space, you can give it the topology of $\mathbb{R}^n$, where $n$ is the dimension. Then any endomorphism (linear map) of this vector space is continuous. If you remove the assumption that the space is finite-dimensional, you can find discontinuous linear maps.