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{Showing that Show $U_n$ is also dense.}

Since by {Second.} with $X = \mathbb{R}$, we know that upon fixing $n,$ that every point of continuity of $f$ belongs to $U_n$ and since $f$ is assumed to be continuous at all rational points, then $ U_n$ contains $\mathbb Q$. And since $\mathbb{Q}$ is dense in $\mathbb{R}$(by Advanced Calculus), hence $U_n$ is dense in $\mathbb{R}.$ \

Where Second. is "Showing that $\bigcap_n U_n$ is precisely the set of points at which $f$ is continuous."

Is this proof correct? I am confused about directly saying that $U_{n}$ is dense in $\mathbb{R}$ because $\mathbb{Q}$ is dense in $\mathbb{R}$. Is there details that should be added?

  • What are the $U_n$? – ViktorStein Mar 19 '20 at 02:39
  • @ViktorGlombik For $n\in\mathbb{N}$, consider the sets $$ U_n:={x\in X:\exists\delta>0,\forall y,z \in X, , y,z\in B(x,\delta)\implies |f(y) - f(z)|<1/n}. $$ –  Mar 19 '20 at 02:44
  • @ViktorGlombik could you please look at this question if you have time https://math.stackexchange.com/questions/3586277/understanding-why-we-take-g-in-l10-1-setminus-l20-1-and-f-in-b-n-in – Intuition Mar 19 '20 at 02:47

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You could put some more details into it, for example this way:

If $A\subseteq B$, then $\bar{A}\subseteq\bar{B}$. Now, choose $A=\mathbb{Q}$ and $B=U_n$. Since $\bar{\mathbb{Q}}=\mathbb{R}$, it follows that $\mathbb{R}\subseteq \bar{U_n}\subseteq\mathbb{R}$.

The idea is correct, though. If a set contains a dense set, then it is dense itself, just as you claimed.

  • are you saying that this claim is not always correct? –  Mar 19 '20 at 00:24
  • and is $U_{n}$ dense or the intersection that is dense? –  Mar 19 '20 at 00:28
  • For $n\in\mathbb{N}$, consider the sets $$ U_n:={x\in X:\exists\delta>0,\forall y,z \in X, , y,z\in B(x,\delta)\implies |f(y) - f(z)|<1/n}. $$ –  Mar 19 '20 at 02:44