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Let's consider the set $X := \{(x,\,0,\,0)\in \mathbb{R^{3}}: 0 < x < 1\}$. Under the usual topology of $\mathbb{R^3}$, is this set open?

My guess it is not, if we sketch it, but how can one analytically prove this, in terms of open balls and? Thanks in advance!

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You can take the point $x_0=(\frac{1}{2},0,0)$ that belongs to $X$ and consider the open ball with center $x_0$ and of ratio $\varepsilon > 0$, then the point $(\frac{1}{2},\frac{\varepsilon}{2},0)$ belongs to the ball but is not in $X$, and this is true for all $\varepsilon > 0$. Then the set is not open.

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Any open ball containing (x,0,0) contains a point (x,y,z) where $y,z \neq 0$

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A one dimensional line (part of) or two dimensional plane (part of) can't contain any 3 dimensional $\epsilon$ ball within and hence no point of which is an interior point and hence they are not open. Your case is a segment of a line.

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