Given $\{(x,y) \in\mathbb{R}^2 | 0 \leq x <1\}$. Show that it is not closed in $(\mathbb{R}^2, \|\cdot\|_2)$ 
Given $\{(x,y) \in\mathbb{R}^2 | 0 \leq x <1\}$. Show that it is not closed in $(\mathbb{R}^2, \|\cdot\|_2)$

The easiest thing I think would be to show that the closure is the set $(x, y)$ where $0 < x < 1$
I imagine you can show that all the points $(1,y)$ are in the closure
Then the only thing that you would have left is to show the set with $(x, y)$ where $0\leq x\leq 1$ is closed
 A: Hint: Build a sequence in the subspace that converges outside of it. For example $a_n=\left(1-\frac1n,0\right)$
A: The easiest way of proving that consists perhaps in finding a point which belongs to the closure but not to the set. Such a point is $(0,1)$, for instance: it doesn't belong to $\{(x,y)\in\mathbb{R}^2\,|\,0\leqslant x<1\}$, but, for every $\varepsilon>0$, if you take $x_0\in(1-\varepsilon,0)$ such that $0\leqslant x_0<1$, then$$(x_0,0)\in D_\varepsilon\bigl((1,0)\bigr)\cap\{(x,y)\in\mathbb{R}^2\,|\,0\leqslant x<1\}.$$
A: You should start with the definition of a "closed" set as a set whose complement is open. So, let
$$S = \left\lbrace \left( x, y \right) \in \mathbb{R}^2 | 0 \leq x < 1 \right\rbrace$$
Then, the complement of this set will be
$$T = \left\lbrace \left( x, y \right) \in \mathbb{R}^2 | x < 0 \text{ or } x \geq 1 \right\rbrace$$
Now, if $S$ would have been closed, then $T$ would have been open. Which means that if $S$ would have been closed, and if we took any point $\left( x_0, y_0 \right) \in T$, then there would have existed a positive real number (let us call it $\epsilon$) such that the open ball of radius $\epsilon$ around the point $\left( x_0, y_0 \right)$ will be completely contained in the set T. Now, open ball of radius $\epsilon$ around the point $\left( x_0, y_0 \right)$.
$$B_{\epsilon} \left( x_0, y_0 \right) = \left\lbrace \left( x, y \right) \in \mathbb{R}^2 | || \left( x, y \right) - \left( x_0, y_0 \right) ||_2 = \sqrt{\left( x - x_0 \right)^2 + \left( y - y_0 \right)^2} < \epsilon \right\rbrace$$
Now, we shall find a point in T which does not satisfy this definition and hence prove that $T$ is not open which further proves that $S$ is not closed.
For the same, consider the point $\left( 1, y \right)$ and make the open ball of radius $\epsilon$, $\epsilon$ being arbitrary.
For every such $\epsilon$, the point $\left( 1 - \dfrac{\epsilon}{2}, y \right) \in B_{\epsilon} \left( 1, y \right)$ but $\left( 1 - \dfrac{\epsilon}{2}, y \right) \notin T$ because $1 - \dfrac{\epsilon}{2} < 1$. Therefore, we conclude that the given set $S$ is not closed.
