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Consider the norms $||x||_1 = \sum_{i=1}^n |x_i|$ and $||x||_2 = \large(\sum_{i=1}^n |x_i|^2 \large) ^{\frac{1}{2}}$ induces the topology $\mathcal T_1 $ and $\mathcal{T_2}$ on $R^n$, the $n-$dimensional euclidean spaces ,then

1.$\mathcal T_1$ is weaker than $\mathcal T_2$

2.$\mathcal T_1$ is stronger than $\mathcal T_2$

3.$\mathcal T_1$ is equivalent to $\mathcal T_2$

4.$\mathcal T_1$ and $ \mathcal T_2$ are incomparable.

Intutively i think (3) is the answer, let x $\in X_1$ and its neighbourhood $B_r^1(x)\ \ in\ \ \mathcal T_1$ , there exist a ball $B_r^1(x) \ \ in \ \ \mathcal T_2$ such that x $\in B_r^2(x) \subseteq B_r^1$ so $\mathcal T_1$ is weaker than $\mathcal T_2$ . how th show that the converse part.

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Actually, the easy part is proving that $T_2$ is weaker than $T_1$. This is due to the inequality $$\tag{1}\lVert y\rVert_2\le \lVert y \rVert_1$$ which holds for all $y\in \mathbb{R}^n$. (Can you prove it? It is an immediate consequence of the subadditivity of $\sqrt{\cdot}$). Specializing (1) to $y=x-x_0$ you see that any ball in $T_1$ contains a ball in $T_2$. – Giuseppe Negro Feb 25 '14 at 9:45
Your intuition is correct. Let $\varepsilon>0$ and $x\in\mathbb{R}^{n}$. Then $x\in B_{\delta}^{2}\left(x\right)\subset B_{\epsilon}^{1}\left(x\right)$ for a $\delta>0$ small enough and also $x\in B_{\delta'}^{1}\left(x\right)\subset B_{\epsilon}^{2}\left(x\right)$ for $\delta'>0$ small enough. Here $B_{\epsilon}^{i}\left(x\right)$ denotes a ball in $\left(\mathbb{R},\mathcal{T}_{i}\right)$ with center $x$ and radius $\epsilon$. – drhab Feb 25 '14 at 9:49

There is a simple but interesting result which says "every two norms on a finite dimensional vector space are equivalent" and then you can conclude the reasoning proving that equivalent norms induce equivalent topologies on the same space.

look here for further details

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True, but I don't think that this answer is satisfactory for the case at hand. It is better to prove directly that the given norms are equivalent. – Giuseppe Negro Feb 25 '14 at 9:42
He can follow the proof and apply it directly to the two norms of the exercise, but I think such a basic argument deserve a generalization – Riccardo Feb 25 '14 at 9:45

$\left\Vert x\right\Vert _{1}<\epsilon\Rightarrow\left\Vert x\right\Vert _{2}<\epsilon\times \sqrt{n}$ (sharper is possible, but there is no need for it)

$\left\Vert x\right\Vert _{2}<\epsilon\Rightarrow\left\Vert x\right\Vert _{1}<\epsilon \times n$ (sharper is possible, but there is no need for it)

This allows you to write a set $U\in\mathcal{T}_{1}$ as a union of balls from $\mathcal{T}_{2}$ and vice versa.

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