Aloizio Macedo
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 1d comment Do monomials form a basis for the vector space of real analytic functions? @user1952009 What is the most natural extension of a concept, if not literally the same concept? May 1 awarded general-topology Apr 30 comment Torus with a point deleted is not a retract of the torus. @JohnHughes A compact set can be a retract of a noncompact set, but not conversely, since continuity preserves compactness (and the question is about the converse situation). But you are right, the remained of the proof applies to the torus minus an open disk : ). Apr 30 comment Torus with a point deleted is not a retract of the torus. @Mark One way to show this is using the fact that $T$ is Hausdorff and connected. It $T \backslash \{x_0\}$ were compact, it would be closed. But it is open, since it is the complement of a point. Since $T$ is connected, then $T \backslash \{x_0\}$ must not be closed. You can also argue directly that it will not be closed, since $x_0$ is clearly a limit point of its complement. Another way is that the torus is a compact metric space, hence complete. If you remove one point, it is not complete anymore (since such a point was clearly not isolated). Therefore, it cannot be compact. Apr 30 revised Torus with a point deleted is not a retract of the torus. added 76 characters in body Apr 30 revised Torus with a point deleted is not a retract of the torus. edited tags Apr 30 answered Torus with a point deleted is not a retract of the torus. Apr 30 comment Compact-open topology on $\operatorname{Hom}_\mathbb{R}(V,W)$ Sorry, I don't understand what you mean by "the topology is given by the vector space structure". What I mean is that, given any two topological vector spaces of the same dimension $X,Y$ (that is, the topology on $X$ and $Y$ are only assumed to be Hausdorff and to make the sum and multiplication by scalar continuous), then any isomorphism between them is a homeomorphism. If you have two different topologies on a vector space $X$, if both of them make $X$ a TVS, then the identity will therefore be a homeomorphism. This fact is a consequence of Theorem $1.21$ of Rudin's FA. Apr 30 comment Compact-open topology on $\operatorname{Hom}_\mathbb{R}(V,W)$ In the case of finite dimension, yes. In finite dimension topological vector spaces, every isomorphism is a homeomorphism. The identity in a vector space is a isomorphism, hence you have your result. If one or both are not finite dimensional, first of all we don't have $\text{Hom}_{\mathbb{R}} (V,W) \subset C^0(V,W)$. Secondly, you must tell what topologies you are considering. However, I don't know any general result. Apr 29 revised How to show that a continous function $f:\mathbb{R}^m \to \mathbb{R}$ has a maximum? added 38 characters in body Apr 29 answered How to show that a continous function $f:\mathbb{R}^m \to \mathbb{R}$ has a maximum? Apr 29 answered Differentiability of vector valued function? Apr 27 answered Argue that the iterated integral of a continuous function is continuous Apr 26 comment $T:R^3 \rightarrow R^3$. Show that there is a line L such that $T(L) =L$. What is the problem? Apr 25 comment Prove $\sum_{n = 0}^{\infty} a_n X^n$ converges at every point in $[-1,1)$ if $a_n$ is non-increasing and converges to $0$ Do you mean $\sum a_n X^n$? Otherwise, what is the radius of convergence of a series of real numbers? Apr 24 revised Every open ball of a normed vector space $E$ its homeomorph to the entire space $E$. added 575 characters in body Apr 24 answered Every open ball of a normed vector space $E$ its homeomorph to the entire space $E$. Apr 24 revised Range of this double trigonometric function edited tags Apr 24 answered Why $\displaystyle\sum_{n=1}^\infty {\ln(n^2) \over (n^2) }$ converges? Apr 24 revised Why does dual basis not span the dual space when the given vector space V is infinite dimensional? edited title