Let $f$ be a continuous  mapping of a metrizable  space $(X,\tau)$  onto a topological space  $(Y,\tau_1)$ .  Is $(Y,\tau_1)$  necessarily metrizable ?
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I noticed this problem on a previous exam that I completely missed, and I was wondering if anyone could help me out.     Suppose $f: Y \rightarrow X$ is a continuous mapping of a separable metric space $Y$ onto a compact Hausdorff space $X$. Show that $X$ is metrizable. Furthermore, give an example of a continuous mapping of a separable metric space onto a mon-metrizable Tychonoff space.   Any help would be greatly appreciated!
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Let $X$ and $Y$ be metric spaces, $X$ compact, and $T:X \to Y$ bijective and continuous. Show that $T$ is a homeomorphism.   My attempt:  We need only show that $T^{-1}$ is continuous.  Let $M \subset Y$ be closed. Then, from the surjectiveness of $T$, there exists a $A \subset X$ such that $T(A)=M$. Since $M$ is closed, by the continuity of $T$, we must have that $A$ is closed.  Now $$T^{-1}(T(A))=T^{-1}(M)$$ i.e, $A=T^{-1}(M)$. Since $A$ and $M$ are closed, we must have that $T^{-1}$ is continuous.  My problem with this, however, is that I am not using the fact that $X$ is compact anywhere in the proof. Is there perhaps another way that I can show this, using the compactness of $X$?   Any hints will be appreciated.
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If a metrizable topological space is totally bounded with one metric, is it totally bounded with all others? A related, stronger question: if every metrization of a topological space is bounded, are they all totally bounded?
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Prove that every continuous image of a separable space is separable.   Let $(X, \mathcal T)$ be a separable space. Then there exists some countable subset $A \subseteq X$ such that $\overline{A}=X$. Let $f:X\to Y$ be a continuous mapping. Notice that $f:X \to f(X)\subseteq Y$ is clearly surjective.  Since $A\subseteq X$, and functions preserve set inclusion, we have that $$f(A) \subseteq f(X)= f(\overline{A}).$$ Also, it is clear that $f(A)$ is itself also countable.  What (I think) I need to show, however, is that $\overline{f(A)}=f(\overline{A})$ (which I am not sure if it's true). This will thus show that $f(A)$ is a countable, dense subset of $f(X).$  Is this the correct approach, or do I have a mistake in my method?  Can anyone please help show me how I can go about finishing this proof?
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Consider the space $\mathcal{K}(\mathbb{R}^{n})$ of compact subsets of $\mathbb{R}^{n}$ endowed with the Hausdorff metric $\rho$, and let $\lambda$ denote the $n$-dimensional Lebesgue measure on $\mathbb{R}^{n}$. Now, I understand that $\lambda$ is not continuous with respect to $\rho$, i.e. that $[ \lim_{k \to \infty} \rho(K, K_{k}) = 0 ] \not \Rightarrow [ \lambda(K) = \lim_{k \to \infty} \lambda(K_{k}) ]$. But my question is: Suppose $\lambda(K_{i}) = \lambda(K_{j}) = C$ for every $i, j \in \mathbb{N}$, i.e. the sequence $k \mapsto \lambda(K_{k})$ is constant; does this imply that $\lambda(\lim_{k \to \infty} K_{k}) = C$? If not, I'm interested to see a counterexample.  Thanks.
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Let $f$ be a continuous mapping of a compact metric space $(X, d)$ onto a Hausdorff space $(Y, \tau_1)$. Then $(Y, \tau_1)$ is compact and metrizable.   In one proof the following metric is constructed:  $d_1(y_1, y_2) = inf\{d(a, b) : a \in f^{-1}\{y_1\},\ b \in f^{-1}\{y_2\}\}$, for all $y_1$ and $y_2$ in $Y$.  I'm thinking about how to prove that triangle inequality holds for $d_1$, s.t. $d_1(x, z) \leq d_1(x, y) + d_1(y, z)$  And how about the case when triangle inequality does not hold?    Update: @daniel-schepler suggested to use metric $d_1(x, y) = inf\{d(x^*, x_1) + d(y_1, x_2) + \cdots + d(y_{n-1}, x_n) + d(y_n, y^*)\}$ where $f(x^*) = x$, $f(y^*) = y$, and $f(x_i) = f(y_i)$ for each $i$.  Now we need to prove 3 property of metric: positivity, symmetry and triangle inequality. This is my sketch:   Positivity. Chose two points $x,y \in Y$ s.t. $x \neq y$. Singleton sets $\{x\}$ and $\{y\}$ are disjoint closed sets by Hausdorfness of $Y$ and their preimages are also disjoint closed in $X$ by continuity of $f$. $X$ is compact metric space by hypothesis and therefore a normal space. In normal space every two disjoint closed sets have disjoint open neighborhoods. Let $f^{-1}(x) \subseteq U$ and $f^{-1}(y) \subseteq V$ s.t. $U,V$ are open and $U \cap V = \varnothing$. Therefore there exists an $\varepsilon &gt; 0$, s.t. $x^* \in B_\varepsilon(x^*) \subseteq U$ and $y^* \in B_\varepsilon(y^*) \subseteq U$ and distance between any two closed sets is strictly positive and $d_1$ is positive and zero iff $x = y$. Symmetry. Clearly $d_1$ is symmetric. Triangle inequality. For triangle inequality we can show that in the "worst" case $d_1(x, z) = d_1(x,y) + d_1(y, z)$.
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The question is (from Topology without tears) that: Let $(X,d)$ be a metric space and $\tau$ the corresponding topology on $X$. Fix $a \in X$. Prove that the map $f:(X,\tau) \rightarrow \mathbb{R}$ defined by $f(x) = d(a,x)$ is continuous.  My first attempt was that let an open ball, $B_{\epsilon}(a)\subset \mathbb{R}$ and show that $f^{-1}(B_{\epsilon}(a)) \subset \tau$. Since $f(x) = d(a,x)$, $f^{-1}(B_{\epsilon}(a))=B_{\epsilon}(a) \in \tau$.   I am pretty sure the last sentence is wrong. I am not sure if every open ball is an open set in topology induced by metric space.  Please give me some hint and direction. Thanks
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Let $(X,d)$ be a metric space. A function $f:X\rightarrow X$ maps Cauchy sequences to Cauchy sequences. Then if $x_n\rightarrow x$, then $\{x_n\}$ is Cauchy, implying $\{f(x_n)\}$ is Cauchy. But if $X$ is not complete then $\{f(x_n)\}$ may not converge, let alone converging to $f(x)$. So the condition doesn't imply $f$ is continuous.   However what about the converse? If $f$ is continuous does it take Cauchy sequences to Cauchy sequences?
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Hello I am doing this exercise where we let X and Y be topological spaces and $f:X\to Y$ a function. We suppose that $\mathcal{A}$ is a collection of closed subsets of X, such that $\bigcup_{A\in\mathcal{A}} A = X$. We assume that for every $A\in \mathcal{A}$ the restriction $f_{|A}: A\to Y$ of f to A is continuous. Furthermore we suppose that for each $x\in X$ there exists a neighborhood U of x such that the intersection $A\cap U$ is non-empty for only finitely many $A\in\mathcal{A}$. And I have to show that f is continuous. I know that since $f_{|A}: A\to Y$ is continuous we have that for all open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of A. And to show that $f: X\to Y$ is continuous I am thinking of showing that for each open subsets $O\subseteq Y$ the inverse image $f^{-1}(O)$ is an open subset of X, but this would be obvious wouldn't it since for all open $O\subseteq Y$ we have $f^{-1}(O)\subseteq A\subseteq X$?
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My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!!
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I was asked the following question: if two topological spaces $X, Y$ are such that there exist a function $f:X\rightarrow Y$ continuos and bijective and a function $g:Y\rightarrow X$ continuous and bijective, then are the two spaces homeomorphic?  I would say that the answer is No, but I didn't find any counterexample so far.
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Let $X$ be the subspace $[0,1] \cup [2,3]$ of $\mathbb{R}$, and let $Y$ be the subspace $[0,2]$ of $\mathbb{R}$. The map $p \colon X \to Y$ defined by  $$ p(x) \colon=  \begin{cases} x \ &amp;\mbox{ for } \ x \in [0,1], \\ x-1 &amp;\mbox{ for } \ x \in [2,3]  \end{cases}  $$ is surjective and continuous. This much is clear to me.   But Munkres asserts that this map is also a closed (and hence a quotient) map. How to verify this?   Now if $A$ is the subspace $[0,1) \cup [2,3]$ of $X$, then the map $q \colon A \to Y$ obtained by restricting $p$ is continuous and surjective. Alright!   Now what stumps me is the following assertion by Munkres:   The set $[2,3]$ is saturated with respect to $q$. How to show this?
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Say that a topological space is CH if it is both compact and Hausdorff.   Let $T$ and $T'$ be two topologies on the same set X that are comparable   but different, i.e., $T$ is either strictly smaller or strictly larger than $T'$.   Assume that $(\text{X}; T )$ is CH. Prove that $(\text{X}; T')$ is not CH.   I just can seem to get started, so I was wondering if anybody could give me a hint...  If $T' \subsetneqq T$, then the identity $id: (\text{X},T) \rightarrow (\text{X}, T')$ is continuous and surjective. This implies that $id(\text{X},T) = (\text{X},T')$ is also compact. So in order to show that $(\text{X},T')$ is not CH, I need to show that it is not Hausdorff, right? But I'm stuck...
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Suppose $f:X\rightarrow Y$ is a continuous map. Define the $\textbf{graph of f}$  to be $\Gamma(f)=$ $\{$ $(x,y)$ $\in X\times Y$ $:$ $y=f(x)$ $\}$ . Then the map $\phi_f: X \rightarrow \Gamma(f)$, given by $\phi_f(x)=(x,f(x))$ is a homeomorphism.  Solution:  Observe that $\phi_f$ is bijective, because it has an inverse, $\phi^{-1}_f : \Gamma(f) \rightarrow X$ $,$ $\phi^{-1}_f(x,f(x))=x$. Hence $\phi_f(X)=\Gamma(f)$. I will consider the function $\phi_f$ from $X$ to $X \times Y$ and show that it is continuous. Since $\pi_1 \circ \phi_f$ $=$ $id_X$ and the identity is continuous, this particular composition is continuous. Since $\pi_2 \circ \phi_f$ $=$ $f$ and $f$ is continuous by assumption, it follows that this composition is continuous and that $\phi_f : X\rightarrow X\times Y$ is continuous. Restricting the codomain  to $\Gamma(f)$ implies that $\phi_f : X\rightarrow \Gamma(f)$ is continuous. Since it is clearly surjective, it follows that it is surjective topological embedding, which is a homeomorphism.  Corollary: Let $U\subseteq \mathbb{R}^n$ be an open subset and $f: U\rightarrow \mathbb{R}^k$ be any continuous function. Then $\Gamma(f) \subseteq \mathbb{R}^{n+k}$ is a manifold.  Proof: Clearly the graph of $f$ is second countable and hausdorff as a subspace. Let $p \in \Gamma(f)$. Then by the above theorem, a homeomorphism between $U$ and $\Gamma(f)$ exists. Hence the graph of $f$ is  manifold.  Please tell me if this is correct first and then include additional comments, please.
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Pugh, "Mathematical Analysis", exercise 2.17:     Assume $f : M \to N$ is a map from one metric space to another which satisfies the following condition: for every convergent sequence $(a_n) \subset M$, the sequence $(f(a_n)) \subset N$ converges. Prove that $f$ is continuous.   (I know that if $f$ commutes with limits then the answer is affirmative, but I can't prove it with that hypothesis).
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Following a reference from "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias.     Since the sum and the product of real continous functions, defined in any topological space, are continuous we can easily prove that every metric space is a Tychonoff spaces. So we suppose that $(X,d)$ is a metric space and we consider a closed set $F$ a point $p\notin F$ and so we define a function $f:X\rightarrow[0,1]$ by the condiction    $$ f(x)=\frac{d(x,p)}{d(x,p)+d(x,F)} $$   where $d(x,F):=\mathscr{inf}\{d(x,a):a\in F\}$. So we observe that $f(p)=0$ and $f(x)=1$ for any $x\in F$; then $f$ is continuous, since any function $g:X\rightarrow\mathbb{R}$ defined as    $$ g(x)=d(x,A) $$       for any $x\in X$ and for any fixed $A\subseteq X$, is a countinuous function and so $f$ is continuous  function, since the sum and the ratio of continuous function is a continuous function.       Well by definition of Tychonoff space we can claim that the metric space $(X,d)$ is a Tychonoff space.   Well unfortunately I don't be able to prove that the function $g$ is continuous: could someone help me, please?
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A continuous function maps a compact set to a compact set. Is the converse of this true? That is, is a function that maps every compact set to a compact set necessarily continuous?