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.
 A: A continous function $h$ can't assume $0$ in $x=1$ and $1$ in $\mathbb{R}-\{1\}$ in fact $h(1)=0 \implies\exists$ a neighborhood $I$ of $1$ such that $h(x) < 1$ for $x \in I$ because $h$ is continous.
A: Without knowing the entirety of the model answer you're describing, here is my best guess at what it is trying to say:
If $h$ is an identity then $g(x)h(x)=g(x)$ for every $g$ and every $x$. In particular this has to be true for $g(x)=x-1$ (which is clearly in $R$).
However, then we have $(x-1)h(x)=x-1$ everywhere, and for every $x\ne 1$ we can cancel the $x-1$. Thus we must have $h(x)=1$ for all $x\ne 1$.
But this means that $\lim_{x\to 1} h(x)=1$, which means that we can't have $h$ be continuous and at the same time $h(1)=0$. No matter which of these fails, this contradicts the assumption that $h$ is an identity within $R$.
A: You say "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 $g(1)=0$."
I deduce that you must be misquoting the answer - you do understand why the function $h(x)=1$ is not in $R$, since $h(1)\ne0$.
I bet the answer actually says this: ($*$)"There does not exist $h\in R$ such that $h(x)=1$ for all $x\ne1$."
I believe that that's what it says because that makes much more sense as an answer, and also because I can believe it's possible for you not to understand that! If in fact the only condition on $g\in R$ was $g(1)=0$ then ($*$) would be false, which would explain why you couldn't understand it. But $g(1)=0$ is not the only condition on $g\in R$! The functions in $R$ are also required to be continuous.
And that's why ($*$) is true: If $h(x)=1$ for all $x\ne0$ and also  $h$ is continuous, then $h(1)=1$, so $h\ne R$.
(See Henning Malcolm's  answer for an explanation of why this means $R$ does not have an identity - I've ignored the actual question here, trying instead to explain what I suspect you're missing, as requested.)
A: *

*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.
A: The multiplication in $R$ is defined by
$$
fg\colon x\mapsto f(x)g(x)
$$
for $f,g\in R$. Suppose an identity exists, call it $e$.
Suppose $x_0\ne1$; then there is $f\in R$ such that $f(x_0)=1$. Indeed, you can consider $f(x)=(x-1)/(x_0-1)$ as this function is continuous and $f(1)=0$. Since $e$ is the identity, we must have $ef=f$, so in particular
$$
e(x_0)f(x_0)=f(x_0)
$$
that entails $e(x_0)=1$. So the function $e$ must satisfy $e(x)=1$ for every $x\ne1$. As a consequence
$$
\lim_{x\to1}e(x)=1
$$
However, $e\in R$, so $e$ is continuous; this implies $e(1)=1$: a contradiction, because $e\in R$ implies $e(1)=0$.
