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If $\Delta = [a, b, c]$ is a triangle having the origin in its interior and $\Delta^{*} = [a, b, c] \setminus \{0\}$ is the same triangule with the origin removed. In topological space $X = \mathbb{R}^{2} \setminus \{0\}$, does $\Delta^{*}$ have to be a simplex?

I believe so, because not being a simplex would mean that there is no continuous function $\sigma: \Delta_{2} \longrightarrow \Delta^{*}$, where

$$\Delta_{2} = \left\{\sum_{i=0}^{2} t_{i} \vec{e}_{i}: \ t_{i} \in [0, 1] \ and \ \sum_{i=0}^{2} t_{i} = 1 \right\},$$

is the standard 2-simplex. But it seems strange to me. Is this reasoning correct?

Appreciate.

José Psicodélico
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2 Answers2

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Since $\Delta^*$ is not compact it can't be a continuous image of the compact space $\Delta_2$.

Christoph
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You're right, $\Delta^*$ is not a simplex. One reason is that a simplex is compact but $\Delta^*$ is not compact.

Your concern about $\sigma_2$ is not an issue. Just because $\Delta^*$ is not a simplex does not mean there are no continuous maps $\sigma_2 : \Delta^2 \to \Delta^*$. There exist a lot of continuous maps like that, for example any constant map whose image is some point of $\Delta^*$.

What doesn't exist is a continuous surjective map $\sigma_2 : \Delta^2 \to \Delta^*$, because $\Delta^2$ is compact, $\sigma_2$ is continuous, the continuous image of a compact set is compact, but $\Delta^*$ is not compact.

Lee Mosher
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    No, $\Delta^$ is not compact. The point $0$ is not an isolated point of the simplex $[a,b,c]$, meaning that one can construct a sequence of points $p_i \in [a,b,c] - {0}$ which converges to $0$. When you remove $0$ to get the subset $\Delta^ = [a,b,c]-{0}$, the sequence $(p_i)$ is in $\Delta^$ but does not converge to anything in $\Delta^$, in fact its subsequences all have nothing to converge to (because all of its subsequences converge to $0$ which has been removed). This proves that $\Delta^*$ is not compact. – Lee Mosher Feb 17 '20 at 19:22