Let $R$ be the ring of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Let $A=\{f \in R \mid f(0)\text{ is an even integer}\}$. Show that $A$ is a subring of $R$, but not an ideal of $R$.

Silly doubt

Under what binary operations this is a ring? Usually in the case of group it is composition. How to proceed here?

If that part is clear, then this problem can be solved.


The operations are addition and multiplication of functions, i.e. $$ (f+g)(x)=f(x)+g(x) \\ (fg)(x)=f(x)g(x). $$

  • $\begingroup$ Thanks for answering. So if nothing is given we assume this ? $\endgroup$ – blue boy Sep 10 '18 at 12:50
  • $\begingroup$ @blueboy Yes, you do. $\endgroup$ – Janik Sep 10 '18 at 12:53

The implicit multiplication of the ring is pointwise multiplication of functions, so you only need to find a continuos function that multiplied pointwise with an element of the set takes you out of the set.


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