hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | o =        | r = \left({a + b}\right)^2 + \left({a - b}\right)^2       | c =  }} {{eqn | r = a^2 + 2 a b + b^2 + a^2 - 2 a b + b^2       | c = Square of Sum and Square of Difference }} {{eqn | r = a^2 + b^2 + a^2 + b^2 }} {{eqn | r = 2 \left({a^2 + b^2}\right)       | c =  }} {{end-eqn}} {{qed}}
{{:Euclid:Proposition/II/9}} :400px That is, from the above diagram: : $\left({AC + CD}\right)^2 + \left({AC - CD}\right)^2 = 2 \left({AC^2 + CD^2}\right)$ Let $AB$ be bisected at $C$, and let $D$ be another random point on $AB$. Then the squares on $AD$ and $DB$ are double the squares on $AC$ and $CD$. The proof is as follows. Construct $CE$ perpendicular to $AB$ and construct $CE = AC$. Join $EA$ and $EB$. Construct $DF$ parallel to $CE$ to intersect $BE$ at $F$, and join $AF$. Construct $FG$ parallel to $AB$ to intersect $CE$ at $G$. Since $AC = EC$ we have from Isosceles Triangle has Two Equal Angles that $\angle EAC = \angle AEC$. Since $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles we have that $\angle EAC + \angle AEC$ also equals a right angle. As they are equal, $\angle EAC$ and $\angle AEC$ are both equal to half a right angle. For the same reason, $\angle EBC$ and $\angle BEC$ are also both equal to half a right angle. We have that $\angle GEF = \angle BEC$, so $\angle GEF$ is also half a right angle. From Parallelism implies Equal Corresponding Angles, $\angle EGF$ is a right angle. So from Sum of Angles of Triangle Equals Two Right Angles, $\angle EFG$ is half a right angle. So $\angle GEF = \angle GFE$, and so from Triangle with Two Equal Angles is Isosceles, $EG = GF$. From Parallelism implies Equal Corresponding Angles, $\angle FDB$ is a right angle. So from Sum of Angles of Triangle Equals Two Right Angles, $\angle BFD$ is half a right angle. So $\angle BFD = \angle FBD$, and so from Triangle with Two Equal Angles is Isosceles, $FD = DB$. Since $AC = CE$, the square on $AC$ equals the square on $CE$. So the squares on $AC$ and $CE$ together are twice the square on $AC$. But from Pythagoras's Theorem, the square on $AE$ equals the squares on $AC$ and $CE$ because $\angle ACE$ is a right angle. So the square on $AE$ is double the square on $AC$. By a similar argument, the square on $EF$ is double the square on $GF$. But from Opposite Sides and Angles of Parallelogram are Equal, $GF = CD$. So the square on $EF$ is double the square on $CD$. So the squares on $AE$ and $EF$ are double the squares on $AC$ and $CD$. We have that $\angle AEF$ is a right angle. So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AE$ and $EF$. So the square on $AF$ is double the squares on $AC$ and $CD$. We have that $\angle ADF$ is a right angle. So from Pythagoras's Theorem, the square on $AF$ equals the squares on $AD$ and $DF$. So the squares on $AD$ and $DF$ are double the squares on $AC$ and $CD$. But $DF = DB$. Hence the result. {{Qed}}
{{:Euclid:Proposition/II/10}} :400px That is, from the above diagram: :$\paren {AC + CD}^2 + \paren {CD - AC}^2 = 2 \paren {AC^2 + CD^2}$ Let $AB$ be bisected at $C$, and let $AB$ be produced to some point $D$. Then the squares on $AD$ and $BD$ are double the squares on $AC$ and $CD$. The proof is as follows. Construct $CE$ perpendicular to $AB$ and construct $CE = AC$. Join $EA$ and $EB$. Construct $EF$ parallel to $AB$, and Construct $DF$ parallel to $CE$. From Parallelism implies Supplementary Interior Angles, $\angle CEF + \angle EFD$ equal two right angles. Therefore $\angle FEB + \angle EFD$ is less than two right angles. So by the Parallel Postulate, $EB$ and $FB$ must meet when produced in the direction $B$ and $D$. Let this point be $G$, and let $AG$ be joined. From Isosceles Triangle has Two Equal Angles, as $AC = AE$, $\angle EAC = \angle AEC$. Because $\angle ACE$ is a right angle, from Sum of Angles of Triangle Equals Two Right Angles, each of $\angle EAC$ and $\angle AEC$ are each half a right angle. For the same reason, each of $\angle CEB$ and $\angle EBC$ are each half a right angle. Since $\angle EBC$ is half a right angle, from Two Straight Lines make Equal Opposite Angles, $\angle DBG$ is also half a right angle. From Parallelism implies Equal Alternate Angles, $\angle BDG = \angle DCE$, so $\angle BDG$ is also a right angle. So from Sum of Angles of Triangle Equals Two Right Angles, $\angle DGB$ is half a right angle. As $\angle DGB = \angle DBG$, from Triangle with Two Equal Angles is Isosceles, we have that $BD = BG$. From Opposite Sides and Angles of Parallelogram are Equal, $\angle EFD = \angle ECD$ which is a right angle,  Since $\angle EGF$ is half a right angle, and $\angle EFD$ is a right angle, $\angle FEG$ is half a right angle from Sum of Angles of Triangle Equals Two Right Angles. So $\angle EGF = \angle FEG$, and so from Triangle with Two Equal Angles is Isosceles, $GF = EF$. Since $EC = CA$, the square on $EC$ equals the square on $CA$. So the squares on $EC$ and $CA$ are together twice the square on $CA$. From Pythagoras's Theorem, the square on $EA$ is equal to the the squares on $EC$ and $CA$, and so twice the square on $AC$. Again, since $FG = FE$, the square on $FG$ equals the square on $FE$. So the squares on $FG$ and $EF$ are together twice the square on $FE$. From Pythagoras's Theorem, the square on $EG$ is equal to the the squares on $GF$ and $FE$, and so twice the square on $EF$. From Opposite Sides and Angles of Parallelogram are Equal, $EF = CD$. But the square on $EA$ has been shown to be double the square on $AC$. So the squares on $AE$ and $EG$ are twice the squares on $AC$ and $CD$. From Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AE$ and $EG$. So the square on $AG$is double the squares on $AC$ and $CD$. But by Pythagoras's Theorem, the square on $AG$ is equal to the the squares on $AD$ and $DG$. As $DG = DB$, the result follows. {{qed}}
:250px Let $AB$ be the given straight line segment. Construct the square $ABDC$ on $AB$. Bisect $CA$ at $E$ and join $BE$. Produce $CA$ to $F$ and make $EF = BE$. Construct the square $AFGH$ on $AF$. Produce $GH$ to $K$. Then $H$ is the point at which $AB$ has been cut so as to make the rectangle contained by $AB$ and $BH$ is the same size as the square on $AH$. The proof is as follows. From Square of Sum less Square, the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the square on $EF$. But $EF = EB$, so the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the square on $BE$. By Pythagoras's Theorem, the squares on $AB$ and $AE$ equal the square on $BE$, because $\angle EAB$ is a right angle. So the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the squares on $AB$ and $AE$. Subtract the square $AE$ from each. Then the rectangle contained by $CF$ and $FA$ equals the square on $AB$. Now the rectangle contained by $CF$ and $FA$ is $\Box CFGK$ because $AF = FG$. Also, the square on $AB$ is $\Box ABDC$. So $\Box CFGK = \Box ABDC$. Subtract $AHKC$ from each. Then $\Box FGHA = \Box HBDK$. Now $\Box HBDK$ is the rectangle contained by $AB$ and $BH$, because $AB = BD$. Also, $\Box FGHA$ is the square on $AH$. Hence the result. {{qed}} {{Euclid Note|11|II}}
:250px Let $\triangle ABC$ be an obtuse triangle where $\angle BAC$ is obtuse. Produce $CA$ past $A$. Construct a perpendicular $DB$ from $CA$ produced. Then the square on $BC$ is greater than those on $BA$ and $AC$ by twice the rectangle contained by $CA$ and $AD$. The proof is as follows. From Square of Sum, the square on $DC$ equals the squares on $CA$ and $AD$ plus twice the rectangle contained by $CA$ and $AD$. Add the square on $DB$ to each. So the squares on $CD$ and $DB$ together equal the squares on $DB$, $CA$ and $AD$ plus twice the rectangle contained by $CA$ and $AD$. But by Pythagoras's Theorem, the square on $CB$ equals the squares on $CD$ and $DB$ because $\angle BDC$ is a right angle. Also by Pythagoras's Theorem, the square on $AB$ equals the squares on $AD$ and $DB$ because $\angle BDA$ is a right angle. So the square on $CB$ equals the squares on $CA$ and $AB$ plus twice the rectangle contained by $CA$ and $AD$. Hence the result. {{qed}} {{Euclid Note|12|II|This is essentially a restatement of the Law of Cosines for obtuse triangles.}}
:250px Let $\triangle ABC$ be an acute triangle, and so by definition, $\angle ABC$ is acute. Construct a perpendicular $AD$ from $BC$. Then the square on $AC$ is less than those on $AB$ and $BC$ by twice the rectangle contained by $BC$ and $BD$. The proof is as follows. From Square of Difference, squares on $BC$ and $BD$ equal the square on $DC$ plus twice the rectangle contained by $BC$ and $BD$. Add the square on $AD$ to each. So the squares on $BC$, $BD$ and $AD$ together equal the squares on $DC$ and $AD$ plus twice the rectangle contained by $BC$ and $BD$. But by Pythagoras's Theorem, the square on $AB$ equals the squares on $AD$ and $BD$ because $\angle ADB$ is a right angle. Also by Pythagoras's Theorem, the square on $AC$ equals the squares on $AD$ and $DC$ because $\angle ADC$ is a right angle. So the  squares on $AB$ and $BC$ equals the square on $AC$ plus twice the rectangle contained by $BC$ and $BD$. Hence the result. {{qed}} {{Euclid Note|13|II}}
:500px Let $A$ be the given polygon. Construct the rectangle $BCDE$ equal to the given polygon. If it so happens that $BE = ED$, then $BCDE$ is a square, and the construction is complete. Suppose $BE \ne ED$. Then WLOG suppose $BE > ED$. Produce $BE$ from $E$ and construct on it $EF = ED$. Bisect $BF$ at $G$. Construct the semicircle $BHF$ with center $G$ and radius $GF$ (see diagram). Produce $DE$ from $D$ to $H$. From Difference of Two Squares, the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the square on $GF$. But $GF = GH$. So the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the square on $GH$. From Pythagoras's Theorem, the square on $GH$ equals the squares on $GE$ and $EH$. Then the rectangle contained by $BE$ and $EF$ together with the square on $EG$ is equal to the squares on $GE$ and $EH$. Subtract the square on $GE$ from each. Then the rectangle contained by $BE$ and $EF$ is equal to the square on $EH$. So the square on $EH$ is equal to the rectangle $BCDE$. So the square on $EH$ is equal to the given polygon, as required. {{qed}} {{Euclid Note|14|II}}
:300px Draw any chord $AB$ on the circle in question. Bisect $AB$ at $D$. Construct $CE$ perpendicular to $AB$ at $D$, where $C$ and $E$ are where this perpendicular meets the circle. Bisect $CE$ at $F$. Then $F$ is the center of the circle. The proof is as follows. Suppose $F$ were not the center of the circle, but that $G$ were instead. Join $GA, GB, GD$. As $G$ is (as we have supposed) the center, then $GA = GB$. Also, we have $DA = DB$ as $D$ bisects $AB$. So from Triangle Side-Side-Side Equality: : $\triangle ADG = \triangle BDG$ Hence: : $\angle ADG = \angle BDG$ But from {{EuclidDefLink|I|10|Right Angle}}: :''{{:Definition:Euclid's Definitions - Book I/10 - Right Angle}}'' So $\angle ADG$ is a right angle. But $\angle ADF$ is also a right angle. So $\angle ADG = \angle ADF$, and this can happen only if $G$ lies on $CE$. But then as $G$ is, as we suppose, at the center of the circle, then $GC = GE$. Thus it follows that $G$ bisects $CE$. But then $GC = FC$, and so $G = F$. Hence the result. {{qed}} === Porism === {{:Finding Center of Circle/Porism}}
From Perpendicular Bisector of Chord Passes Through Center, $CE$ passes through the center of the circle. The center must be the point $F$ such that $FE = FC$. That is, $F$ is the bisector of $CE$.
:300px Let $ABC$ be a circle where $A, B$ are two points taken at random on its circumference. Suppose the straight line $AB$ does ''not'' lie entirely within the circle $ABC$. Then it will lie wholly outside (otherwise it will cross the circumference somewhere, and so we can take the part of $AB$ which ''does'' lie outside). Find the center $D$ of circle $ABC$. Draw $DA$, $DB$ and $DE$, and let $F$ be the point where $DE$ crosses the circumference. As $DA = DB$  then from Isosceles Triangle has Two Equal Angles it follows that $\angle DAE = \angle DBE$. Since one side of $AEB$ of $\triangle DAE$ is produced, it follows from External Angle of Triangle Greater than Internal Opposite that $\angle DEB$ is greater than $\angle DAE$. But $\angle DAE = \angle DBE$, so $\angle DEB$ is greater than $\angle DBE$. But from Greater Angle of Triangle Subtended by Greater Side, $DB$ is greater than $DE$. But $DB = DF$, so $DF$ is greater than $DE$. But $DF$ is less than $DE$. Thus we have a contradiction. So the straight line $AB$ does not fall outside the circle. Similarly, we can show that $AB$ does not lie on the circumference itself. Hence the result. {{qed}} {{Euclid Note|2|III}}
:300px Let $F$ be the center of the circle in question. Draw any chord $AB$ on the circle. Bisect $AB$ at $D$. Construct $CE$ perpendicular to $AB$ at $D$, where $D$ and $E$ are where this perpendicular meets the circle. Then the center $F$ lies on $CE$. The proof is as follows. Join $FA, FD, FB$. As $F$ is the center, $FA = FB$. Also, as $D$ bisects $AB$, we have $DA = DB$. As $FD$ is common, then from Triangle Side-Side-Side Equality, $\triangle ADF = \triangle BDF$. In particular, $\angle ADF = \angle BDF$; both are right angles. From {{EuclidDefLink|I|10|Right Angle}}: :''{{:Definition:Euclid's Definitions - Book I/10 - Right Angle}}'' So $\angle ADF$ and $\angle BDF$ are both right angles. Thus, by definition, $F$ lies on the perpendicular bisector of $AB$. Hence the result. {{qed}}
:300px Let $ABC$ be a circle, in which $AB$ is a chord which is not a diameter (i.e. it does not pass through the center). === First part === Let $CD$ be a diameter which bisects $AB$ at the point $F$. Find the center $E$ of circle $ABC$, and join $EA$ and $EB$. Because $E$ is the center, $EA = EB$. Because $F$ bisects $AB$, $FA = FB$. Also, $FE$ is common, so from Triangle Side-Side-Side Equality $\triangle AFE = \triangle BFE$. From {{EuclidDefLink|I|10|Right Angle}}: :''{{Definition:Euclid's Definitions - Book I/10 - Right Angle}}'' So $\angle AFE$ and $\angle BFE$ are both right angles. Therefore the diameter $CD$ cuts $AB$ at right angles. {{Qed}} === Second part === Let $CD$ be a diameter which cuts $AB$ at right angles at point $F$. We using the same construction as above. Because $E$ is the center, $EA = EB$. From Isosceles Triangle has Two Equal Angles, $\angle EAF = \angle EBF$. But $\angle AFE = \angle BFE$ because both are right angles. As $EF$ is common, it follows from Triangle Side-Angle-Angle Equality that $\triangle AFE = \triangle BFE$. Therefore $AF = FB$. Hence the result. {{qed}} {{Euclid Note|3|III|It is the converse of the Porism to Proposition $1$: Perpendicular Bisector of Chord Passes Through Center.}}
:250px Let $ABCD$ be a circle, in which $AC$ and $BD$ are chords which are not diameters (i.e. they do not pass through the center). Let $AC$ and $BD$ intersect at $E$. {{AimForCont}} they were able to bisect one another, such that $AE = EC$ and $BE = ED$. Find the center $F$ of the circle, and join $FE$. From Conditions for Diameter to be Perpendicular Bisector, as $FE$ bisects $AC$, then it cuts it at right angles. So $\angle FEA$ is a right angle. Similarly, $\angle FEB$ is a right angle. So $\angle FEA = \angle FEB$, and they are clearly unequal. From this contradiction, it follows that $AC$ and $BD$ can not bisect each other. {{qed}} {{Euclid Note|4|III}}
:250px Let $ABC$ and $BDCG$ be circles which cut one another at $B$ and $C$. Suppose they had the same center $E$. Join $EC$ and let $EG$ be drawn at random through $F$. As $E$ is the center of $ABC$, by {{EuclidDefLink|I|15|Circle}}, we have that $EC = EF$. Similarly, as $E$ is also the center of $BDCG$, we have that $EC = EG$. But they are clearly unequal by the method of construction. So from this contradiction, the two circles can not have the same center. {{qed}} {{Euclid Note|5|III}}
If the two circles are outside one another, the result is trivial. This proof will focus on the situation where one circle is inside the other one. :250px Let $ABC$ and $CDE$ be circles which touch one another at $C$, such that $CDE$ is inside $ABC$ {{AimForCont}} they had the same center $F$. Join $FC$ and let $FB$ be drawn at random through $E$. As $F$ is the center of $ABC$, by {{EuclidDefLink|I|15|Circle}}, we have that $FB = FC$. Similarly, as $F$ is also the center of $CDE$, we have that $FC = FE$. But they are clearly unequal by the method of construction. So from this contradiction, the two circles can not have the same center. {{qed}} {{Euclid Note|6|III}}
This follows directly from: : Intersecting Circles have Different Centers : Touching Circles have Different Centers. {{qed}} Category:Circles gwcv49su5996mg850n17qebxym4m3e3
:250px Let $ABCE$ be a circle, and let $AE$ be a diameter of it. Let $D$ be the center of the circle, and let $F$ be a point on $AE$ different from $D$. Let $FB, FC, FG$ be drawn from $F$ to the edge of the circle. Then $FA$ is the greatest, $FE$ is the least, and (according to the diagram above) of the others, $FB$ is greater than $FC$, while $FC$ is greater than $FG$. The proof is as follows. Join $BD, CD, GD$. From Sum of Two Sides of Triangle Greater than Third Side, we have that $DB + DF > BF$. But $AD = BD$, so $AF > BF$. Since $BD = CD$ and $FD$ is common, $BD + DF = CD + CF$. But $\angle BDF > \angle CDF$, so from the Hinge Theorem $BF > CF$. For the same reason, $CF > GF$. Again, since $GF + FD > DG$, and $DG = DE$, $GF + FD > DE$. Subtract $FD$ from each, and we see that $GF > FE$. Therefore $FA$ is the greatest, followed by, in order of size, $FB, FC, FG$ and finally $FE$ is the smallest. ---- Also, from the point $F$ only two equal straight lines fall on the circle $ABCE$, one on either side of the least line $FE$. Construct the angle $\angle FDH$ equal to $\angle GDF$ on the opposite side of $DF$, and join $FH$. Since $GD = DH$ and $DF$ is common, and $\angle FDH = \angle GDF$ from Triangle Side-Angle-Side Equality, $\triangle FDH = \triangle GDF$, and so $FG = FH$. Another straight line equal to $FG$ will not fall on the circle from $F$. For if this is possible, let $FK$ be that straight line. Since $FK = FG$, and $FH = FG$, it follows that $FK = FH$. But this contradicts what we proved earlier. Therefore only one such straight line can be constructed. {{Qed}} {{Euclid Note|7|III}}
We start by establishing a lemma: === Lemma 1 === Let $t \in \R$ be such that $t > 1$, and let $n \in \N$ be such that $n > \dfrac {b - 1} {t - 1}$. Then $b^{\frac 1 n} < t$. <section title="Proof of Lemma 1" foldable="true" header="3"> Because $b > 1$, we have that $b^{\frac 1 n} > 1$. From Sum of Geometric Sequence and this observation, we find that: :$\displaystyle \frac {\paren {b^{\frac 1 n} }^n - 1} {b^{\frac 1 n} - 1} = \sum_{k \mathop = 0}^{n - 1} \paren {b^{\frac 1 n} }^k > n$ Rewriting this inequality, we obtain: :$b - 1 > n \paren {b^{\frac 1 n} - 1}$ As $n > \dfrac {b - 1} {t - 1}$, this means that: :$t - 1 > b^{\frac 1 n} - 1$ We conclude that: :$b^{\frac 1 n} < t$ </section>{{qed|lemma}} Now let $w \in \R$ be any number satisfying $b^w < y$. Then $t = b^{-w} y > 1$, so we can apply the lemma 1. It follows that for $n > \dfrac {b - 1} {t - 1}$: :$b^{\frac 1 n} < b^{-w} y$ and thus: :$b^{w + \frac 1 n} < y$ Similarly, when $b^w > y$ and $n > \dfrac {b - 1} {t - 1}$, we have: :$b^{w - \frac 1 n} > y$ Next, let $W = \set {w \in \R: b^w < y}$. Let $x = \sup W$. The following lemma helps us to verify existence of $x$. === Lemma 2 === For any real number $z > 0$, we have $n \in \N$ such that: :$b^{-n} < z < b^n$ <section title="Proof of Lemma 2" foldable="true" header="3"> Observe that, as $z > 0$, $b^{-n} < z$ whenever $b^n > \dfrac 1 z$. Therefore, it suffices to show that there is an $n \in \N$ such that $b^n > z$ exists. By Sum of Geometric Sequence, we obtain for any $n \in \N$: :$\displaystyle \paren {b - 1}\right) \sum_{k \mathop = 0}^{n - 1} b^k = b^n - 1$ As $b > 1$, we obtain: :$n \paren {b - 1} < b^n - 1$ This implies that any $n$ with $n \paren {b - 1} > z - 1$ has the desired property. </section>{{qed|lemma}} Lemma 2 shows that $x$ is the supremum of a non-empty set with an upper bound, and therefore it exists. We will now prove that $b^x = y$. Suppose $b^x < y$. We have seen that there exists an $n \in \N$ such that $b^{x + \frac 1 n} < y$. That is: :$x + \dfrac 1 n \in W$ But this is impossible as $x = \sup W$. Thus: :$b^x \ge y$ Now suppose $b^x > y$. We have seen that there exists an $n \in \N$ such that $b^{x - \frac 1 n} > y$. That is: :$x - \dfrac 1 n \notin W$ As $x - \frac 1 n$ is not the supremum of $W$, we find a $w \in W$ such that: :$x - \dfrac 1 n < w < x$ But then we have, as $b > 1$: :$b^{x - \frac 1 n} - b^w = b^{x - \frac 1 n} \paren {1 - b^{w - x + \frac 1 n} } < 0$ This is contradicts our initial choice of $x - \dfrac 1 n$ outside $W$. We conclude that $b^x = y$. Lastly, suppose that $b^x = y = b^z$. Then: :$b^{z-x} = \dfrac {b^z} {b^x} = 1 = \dfrac {b^x} {b^z} = b^{x-z}$ It is concluded that $x = z$. {{qed}} {{proofread}} Category:Logarithms 7yax1bhlu5uj784jlv9vux7egyt3rrl
Let $U_n$ be a countable set of open sets of $M$ all of which are everywhere dense. The strategy of this proof is to show that the intersection $\displaystyle \bigcap U_n$ is everywhere dense. Let $W \subseteq A$ be a nonempty open set of $M$. From Open Set Characterization of Denseness, since $U_1$ is everywhere dense: :$W \cap U_1 \ne \O$ Thus, there is a point $x_1 \in A$ and $\epsilon_1 \in R_{>0}$ such that: :$\map {\overline B} {x_1, \epsilon_1} \subset W \cap U_1$ where $\map B {x, \epsilon}$ and $\map {\overline B} {x, \epsilon}$ denote an open $\epsilon$-ball of $x$ and its closure, respectively. Because $U_n$ are everywhere dense, in a recursive manner, we find a pair of sequences $\sequence {x_n}$ and $\sequence {\epsilon_n} > 0$ such that: :$\map {\overline B} {x_n, \epsilon_n} \subset \map B {x_{n - 1}, \epsilon_{n - 1} } \cap U_n$ as well as $\epsilon_n < 1/n $. Since $x_n \in \map B {x_m, \epsilon_m}$ when $n > m$, we have that $x_n$ is a Cauchy Sequence. Thus $x_n$ converges to some limit $x$ by definition of completeness. For any $n$, by closedness: :$x \in \map {\overline B} {x_{n + 1}, \epsilon_{n + 1} } \subset \map B {x_n, \epsilon_n}$ Thus: :$\forall n \in \N: x \in W \cap U_n$ From Open Set Characterization of Denseness it follows that $\displaystyle \bigcap U_n$ is (everywhere) dense. So, by definition, $\struct {M, d}$ is a Baire space. {{qed}} {{Namedfor|René-Louis Baire|cat = Baire}}