Show that this ring has no identity. 
Let $R=\left \{ g:\Bbb{R}\to \Bbb{R} \mid g \text{ is continuous and } g(1)=0 \right \}$ be a ring. Show that $R$ has no identity.

The answer says there does not exist a function $h(x)\in R$ such that $h(x)=1$, which I don't understand why, since the only condition in $R$ is $g(1)=0$. Please help me understand what I am missing.

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*The most obvious choice of identity function is the constant function $h(x) = 1$. After all, multiplying $h(x)$ pointwise by another function $g(x)$ results in $g(x)$ again.

*Unfortunately, $h(x)$ doesn't belong to the ring $R$ because $h(1)=1\neq 0$.  It can't be the identity because it isn't in $R$.

*We can try to repair the problem by defining a different identity function $H(x)$ which is equal to 1 at most points, except $H(1)=0$ as required.

*But then by continuity, $H$ is equal to zero at 1, $H$ is equal to 1 somewhere else, and so at some point $p$, $H(p)=\frac{1}{2}$ (an intermediate value). This is a disaster— take $g(x) = |x-1|$, for example: At that point $p$, $$g(p)H(p) = \frac{1}{2}g(p) \neq g(p).$$  So no such identity  function $H(x)$ exists.
