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I was given the problem to show that for the metric space $(\mathbb{R}^2,d)$ with $d$ the Euclidean distance, the set $D = \left\{(x,x)\in \mathbb{R}^2: x \in [0,1] \right\}$ is closed. This was easy enough, since I can show that for any point $(a,a)$ on this line, any open ball with radius $\varepsilon>0$ around $(a,a)$ would have points outside of $D$ (say for example the point ($-\varepsilon/\sqrt{2}+a,\varepsilon/\sqrt{2}+a$), this is $\varepsilon$ distance away from $(a,a)$ and clearly is not an element of $D$). This showcases every point on $D$ is a boundary point of $D$, and the boundary of any set is closed, hence $D$ is closed.

I then went to change the set by changing the possible values of $x$, hence I got the set $E =\left\{(x,x)\in \mathbb{R}^2: x \in (0,1] \right\}$. This shouldn't be a closed set, since the point $(0,0)$ is a limit point in $E$ (take the sequence $((1/n, 1/n))_{n\geq 1} \in E$). However, I get confused since the "open ball around $(a,a)$ method" I did for $D$ seems fine in this case too, since I can give the same arguments as in the case $D$. Can anyone give me a good explanation on why this method has to fail for the set $E$?

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open ball around the origin, regarding set $E$, yes the same argument shows that the origin is a boundary point for $E$. The origin is also a limit point for $E$, as you said. Since $(0,0) \not\in E$, that is it. $E$ is not closed for that reason.

I think your trouble is you have a method that works in the special case when $M=\partial M$, e.g. when $M=D.$ But $E\not=\partial E$, so you need a different approach for set $E$.

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  • $\begingroup$ $\partial M$ means "the boundary of $M$" $\endgroup$
    – 311411
    Commented Nov 23, 2021 at 23:09

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