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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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    $\begingroup$ What are the operations on $\;R\;$ ? Usual sum and product of functions? And if so, observe that $\;h(1)=1\neq0\;$ , so then $\;h\notin R\;$ ... $\endgroup$ – DonAntonio Apr 7 '18 at 19:18
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    $\begingroup$ The constant function $1$ does not have a root at $x=1$, and is therefore not in $R$. $\endgroup$ – Dave Apr 7 '18 at 19:21
  • $\begingroup$ I suspect you've misquoted the answer. What matters, what I suspect the answer says, and what you might be confused about, is this: There does not exist $h\in R$ such that $h(x)=1$ for every $x\ne1$. $\endgroup$ – David C. Ullrich Apr 7 '18 at 19:22
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    $\begingroup$ @Dave But it's not quite that trivial. Because the constant function $1$ is not the only possible identity - in fact if $h(x)=1$ for every $x\ne1$ then $h$ would be an identity in $R$. $\endgroup$ – David C. Ullrich Apr 7 '18 at 19:24
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    $\begingroup$ @Dave Of course the continuity excludes this! I didn't say the problem was not trivial, just that it's not that trivial. A correct answer has to at least mention continuity: If $S$ is the set of all functions $g$, continuous or not, with $g(1)=0$ then $S$ does have an identity, even though the constant function $h=1$ is not in $S$. $\endgroup$ – David C. Ullrich Apr 7 '18 at 19:35
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  1. 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.

  2. 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 $R$.

  3. 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.

  4. 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.

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  • $\begingroup$ Your #2 is what makes me understand the answer the best. Thank you! $\endgroup$ – numericalorange Apr 7 '18 at 20:16
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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.

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  • $\begingroup$ I'm sorry. I don't get your solution. $1 \in I$, but $g(1) = 0$, so how can you say that $g(x) < 1$ for all $x \in I$? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 19:21
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    $\begingroup$ +1: Thanks for explaining. My bad. I confused it with another property of continuous function. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 19:28
  • $\begingroup$ Thank you for your reply. It is very advanced for me, but thanks! $\endgroup$ – numericalorange Apr 7 '18 at 20:14
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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 $R$.

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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.)

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    $\begingroup$ Oh yes, I did misquote the answer, I apologize. Thank you for taking the time to explain to me what was missing. :) $\endgroup$ – numericalorange Apr 7 '18 at 20:16
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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 $e(1)=0$.

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