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Take for instance $\mathbb{R}^{\mathbb{R}}$ with the product topology or the weak topology on an infinite-dimensional Banach space. You will find more examples here.
We can simplify life a little by dealing with the subspace $S = L - \{l\}$ (for any $l \in L$) instead. So the question becomes if $S$ is a subspace, is $\bar{S}$ a subspace? The key facts are continuity of scalar multiplication and addition. Suppose $x \in \bar{S}$ and $\lambda \neq 0$. Suppose $U$ is open with $\lambda x \in U$. Then ${1 \over \lambda ... 1 It will be compact, as the function that sends$\alpha \in [0,1]$to$\alpha x + (1-\alpha)y$is continuous (as all vector operations are, details omitted). And the continuous image of a compact space ($[0,1]$) is compact. Compact need not imply closed for non-Hausdorff spaces. To give a rather extreme counterexample, suppose that$E$has the indiscrete ... 1 If for every open neighborhood$U$we have$x\in M+U$then$(x-U)\cap M\neq \emptyset $hence$x\in\overline{M}.$1 Let$h\in B(f,\epsilon_0,...,\epsilon_r)\cap B(g,\mu_0,...,\mu_s)$and suppose$r\leq s$. Let$h$be a function in the intersection. We shall write$||q||=\sup_{x\in [a,b]} q(x)$for the sup norm. For$i\leq r$let$\eta_i$be such that$||h^{(i)}-f^{(i)}||+\eta_i<\epsilon_i,||h^{(i)}-g^{(i)}||<\mu_i$. For$i>r,$just let$\eta_i$be such that ... 1 I would have said that the suspension was$X \times [-1, 1]$modulo the relation that $$(x, a) \sim (x', a')$$ if and only iff$a = a' = 1$or$a = a' = -1$, or$a = a'$and$x = x'$. Wikipedia seems to agree with me. It looks as if your author was a little glib, and failed to mention that the "bottom" and "top" sets of equivalent points were not ... 1 In fact in my experience the unreduced suspension is commonly written in the way you did it, although it is wrong. What it means is: collapse one side to a point and also the other side. But NOT both to the same point. So this would be only right when quotiening is associative, i.e. $$SX = (X \times I/ X\times \{0\} )/ (X \times \{1\}) "=" X \times I / ... 1 The topology is generated by sets of the form$$ U_{f,r}=\{g\in C[0,1]: d(\,f,g)<r\}=f+\{g\in C[0,1]: d(0,g)<r\}. $$Note that$$ U_{0,r}+U_{0,r}\subset U_{0,2r}. $$It suffices to show that$$ (\lambda,f)\longmapsto \lambda f,\quad\text{and}\quad (f,g)\longmapsto f+g,$$are continuous. with respect to above topology. a. If$f+g\in V$,$V\subset ...