Open Set and Interior Point

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Recall:

$U$ is open if and only if for any $x\in U$ there is an $\varepsilon >0$ such that $\left( x-\varepsilon ,x+\varepsilon \right)\subseteq U$
Let $X\subseteq\mathbb{R}$. We say that an element $x$ of $X$ is an interior point of $X$ if there is an $\varepsilon >0$ such that $\left( x-\varepsilon ,x+\varepsilon \right)\subseteq X$.

My question is: 1) Can we say that if $U$ is open then for any element $u\in U$ is an interior point?

2) Also, can we say if $U$ has interior points then $U$ is open?

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3 Answers

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Yes, because if $u\in U$ then, since $U$ is open, then, by definition of open set, there is a $\varepsilon>0$ such that $(u-\varepsilon,u+\varepsilon)\subset U$. And this means that $u$ is an interior point of $U$.
No. Take $U=[-1,1]$. Then $U$ is not open (because there is no $\varepsilon>0$ such that $(1-\varepsilon,1+\varepsilon)\subset U$), but $0$ is an interior point of $U$.

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(1) Yes; (2) No.

It is true in general that a set is open if and only if it contains only interior points.

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(1) Yes by definition.

(2) No, since $U$ being an open set requires all $u \in U$ to be interior points of $U$.

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