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I am stuck in this, I have no idea why!

$[0,1]$ is a manifold with boundary, how to justify? Which are the charts?

And how about $[0,1]^2?$ Why it is not a manifold?

My definition of topological manifold with or without boundary is the usual.

- If the sequence satisfies the property lim$_{n\to \infty}(a_n-a_{n-2})=0$, prove that lim$_{n\to \infty}\frac{a_n-a_{n-1}}{n}=0$.
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- $f$ strictly increasing does not imply $f'>0$
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- Does a nonlinear additive function on R imply a Hamel basis of R?

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- Existence of a dense proper subset with non-empty interior?
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- Show that if $f^{-1}((\alpha, \infty))$ is open for any $\alpha \in \mathbb{R}$, then $f$ is lower-semicontinuous.
- show $\exists\ m\in\mathbb{N} \text{ such that: } \quad a+\dfrac{m}{2^n}\geq b$
- dense in terms of order and in terms of the order topology
- Show $\Bbb R ^n$ with a finite set removed is still connected.
- Uncountable Cartesian product of closed interval

But it **is** a manifold with boundary. More precisely, $[0,1]^2$ is a topological space, and as a topological space it is homeomorphic to $D^2 = \{ x \in \mathbb{R}^2 \mid \|x\| \le 1 \}$. You probably already know that $D^2$ can be equipped with charts that make it into a smooth manifold with boundary, and you can transport these charts through the homeomorphism to make $[0,1]^2$ into a smooth manifold with boundary.

However, $[0,1]^2 \subset \mathbb{R}^2$ is not a smooth embedded **sub**manifold with boundary. Whatever smooth charts you will put on $[0,1]^2$ to make it into a submanifold, the inclusion $i : [0,1]^2 \to \mathbb{R}^2$ will not be an embedding.

The problem is at the corners, say $(0,0)$. If $i$ were smooth, then the restriction of $i$ to the boundary would be smooth too. From the topology, we know what the boundary would be: when one of the coordinates is $0$ or $1$.

But then, the differential of $i$ when restricted to $\{0\} \times (0,\epsilon)$ would be proportional to a vertical tangent vector in the tangent space of $\mathbb{R}^2$, whereas the differential when restricted to $(0,\epsilon) \times \{0\}$ would be proportional to a horizontal vector. Since the differential has to be continuous, this is a contradiction as the differential at $(0,0)$ would have be at the same time horizontal and vertical, hence would be the zero vector; but an embedding has to be an immersion.

A manifold with boundary has charts to a half-space. However, near the corners of $I^2$, the natural charts go to a corner space due to the corners. So it is a topological 2-manifold with boundary (after all, it is homeomorphic to the closed disk), but not a smooth 2-manifold with boundary.

Another convenient way to spot it is that the boundary of a manifold with boundary has to be a manifold (without boundary!), but the boundary of $I^2$ has “bad points,” which may not be smoothly parameterized.

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