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